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== 5.2#h(0.6em)Solving Equations


    
      
=== Introduction


      Now that we understand that trigonometric functions can be used to model rotations and cyclical behavior, we can begin to consider some applications. An important part of any model involves solving equations. In Chapter 4 we solved simple trigonometric equations, and now we are equipped to tackle more challenging ones.


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Algebra Refresher]


        It is important to distinguish between an algebraic #emph[expression] and an #emph[equation].  An #strong[equation] is a statement that two algebraic expressions are equal.  It may be true or false, depending on the values of any variables involved.  Here are some examples of equations.


        #math.equation(block: true, alt: "5 open parenthesis 2 plus 6 close parenthesis, equals 5 open parenthesis 2 close parenthesis plus 5 open parenthesis 6 close parenthesis; the square root of 3 squared plus 4 squared, equals 3 plus 4; x squared plus 3 x, equals 10")[$5 ( 2 + 6 ) & = 5 ( 2 ) + 5 ( 6 ) \ sqrt(3^(2) + 4^(2)) & = 3 + 4 \ x^(2) + 3 x & = 10$]
        The first equation is true, the second is false, and the third equation is true only if #math.equation(block: false, alt: "x equals 2")[$x = 2$] or #math.equation(block: false, alt: "x equals minus 5")[$x = − 5$].  When you #strong[solve] an equation, you are finding the values of the variable that make the equation true.
]


      You already know several algebraic techniques for solving equations of different types. But sometimes simple trial and error is the best approach.


      #examplebox("Example 1")[][
+ Evaluate #math.equation(block: false, alt: "2 x plus the cube root of x minus 1")[$" " " " 2 x + root(3, x − 1)$] for #math.equation(block: false, alt: "x equals 9")[$x = 9$].
+ Solve #math.equation(block: false, alt: "2 x plus the cube root of x minus 1 equals 20")[$" " " " 2 x + root(3, x − 1) = 20$].


        
        #solutionbox[
+ Substitute #math.equation(block: false, alt: "x equals 9")[$x = 9$] into the expression to find
        #math.equation(block: true, alt: "2 x plus the cube root of x minus 1, equals 2 open parenthesis 9 close parenthesis plus the cube root of 9 minus 1; equals 18 plus the cube root of 8 equals 20")[$2 x + root(3, x − 1) & = 2 ( 9 ) + root(3, 9 − 1) \  & = 18 + root(3, 8) = 20$]
+ We must find a value for #math.equation(block: false, alt: "x")[$x$] that makes #math.equation(block: false, alt: "2 x plus the cube root of x minus 1")[$2 x + root(3, x − 1)$] equal to #math.equation(block: false, alt: "20")[$20$]. In part (a) we saw that this expression equals #math.equation(block: false, alt: "20")[$20$] when #math.equation(block: false, alt: "x equals 9")[$x = 9$], so the solution of the equation #math.equation(block: false, alt: "2 x plus the cube root of x minus 1 equals 20")[$" " 2 x + root(3, x − 1) = 20 " "$] is #math.equation(block: false, alt: "x equals 9")[$x = 9$].
]
]


      
        Use trial and error to find a solution of the equation


        #math.equation(block: true, alt: "2 x cubed minus 4 x equals 5 x squared minus 3")[$2 x^(3) − 4 x = 5 x^(2) − 3$]
        
          Try small integer values for #math.equation(block: false, alt: "x")[$x$].


        
        
          #math.equation(block: true, alt: "x equals 3")[$x = 3$]


        
      
      Another useful equation-solving method uses graphs.


      #examplebox("Example 2")[][
Use a graph to solve the equation #math.equation(block: false, alt: "x cubed minus 2 x squared minus 5 x equals minus 6")[$" " x^(3) − 2 x^(2) − 5 x = − 6$].


        
        #solutionbox[
We graph the expressions on either side of the equation, that is, we graph #math.equation(block: false, alt: "y equals x cubed minus 2 x squared minus 5 x")[$" " y = x^(3) − 2 x^(2) − 5 x " "$] and #math.equation(block: false, alt: "y equals minus 6")[$y = − 6$]  on the same grid, as shown below.


          We are looking for any values of #math.equation(block: false, alt: "x")[$x$] where the two #math.equation(block: false, alt: "y")[$y$]-values are equal, and these occur at the intersection points of the two graphs. At those points, the #math.equation(block: false, alt: "x")[$x$]-values are #math.equation(block: false, alt: "x equals minus 2 , x equals 1")[$x = − 2 , " " x = 1$] and #math.equation(block: false, alt: "x equals 3")[$x = 3$], and these are the solutions of the equation. You can check that all three values make the equation true.


          #figure(figph[cubic graph], alt: "cubic graph", caption: none)
]
]


      
        Use a graph to show that the equation #math.equation(block: false, alt: "x squared minus 2 x plus 4 equals 0")[$" " x^(2) − 2 x + 4 = 0 " "$] has no real-valued solutions.


        
          The graph does not cross the line #math.equation(block: false, alt: "y equals 0")[$y = 0$].


          #figure(figph[parabola], alt: "parabola", caption: none)


        
      
    
    
      
=== Trigonometric Equations


      The first Ferris wheel was built for the Chicago World's Fair in 1893. It had a diameter of 250 feet and could carry 2160 people in 36 carriages. From the top of the wheel, passengers could see into four states. After loading all the passengers, the wheel made one revolution in nine minutes.


      #figure(figph[Figure], alt: "Figure", caption: none)


      If you are in the bottom carriage of the Ferris wheel at the start of its revolution, your height after #math.equation(block: false, alt: "t")[$t$] seconds is given by


      #math.equation(block: true, alt: "h equals f open parenthesis t close parenthesis equals 139 minus 125 cos ⁡ open parenthesis the fraction 2 t over 3 close parenthesis")[$h = f ( t ) = 139 − 125 cos ( display(frac(2 t, 3)) )$]
      For how long are you more than 240 feet above the ground?


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
In case you are wondering, the boarding platform is 14 feet above the ground, so that #math.equation(block: false, alt: "h equals 14")[$h = 14$] when #math.equation(block: false, alt: "t equals 0")[$t = 0$]. And notice that #math.equation(block: false, alt: "the fraction 2 t over 3 equals 360 degrees")[$display(frac(2 t, 3)) = 360^(∘)$] when #math.equation(block: false, alt: "t equals 540")[$t = 540$] seconds, or 9 minutes.
]


      The figure below shows a graph of the height function and a horizontal line at #math.equation(block: false, alt: "h equals 240")[$h = 240$].


      #figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none)


      From the graph, we see that #math.equation(block: false, alt: "h equals 240")[$h = 240$] at approximately 215 seconds (or 3 minutes and 35 seconds) and 325 seconds (or 5 minutes and 25 seconds) into the ride. Your height is more than 240 feet between those two times, or for about 110 seconds.


    
    
      
=== Solving Trigonometric Equations


      In the example above, we used a graph to solve the equation #math.equation(block: false, alt: "h equals 240")[$h = 240$], or


      #math.equation(block: true, alt: "139 minus 125 cos ⁡ open parenthesis the fraction 2 t over 3 close parenthesis equals 240")[$139 − 125 cos ( display(frac(2 t, 3)) ) = 240$]
      To find a more precise solution, we can use algebraic methods. To start, we'll solve the slightly simpler equation


      #math.equation(block: true, alt: "139 minus 125 cos ⁡ open parenthesis θ close parenthesis equals 240")[$139 − 125 cos ( θ ) = 240$]
      We'll look for all solutions for #math.equation(block: false, alt: "θ")[$θ$] between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "360 degrees")[$360^(∘)$]. We begin by isolating the trigonometric ratio on one side of the equation.


      #math.equation(block: true, alt: "139 minus 125 cos ⁡ open parenthesis θ close parenthesis, equals 240, Subtract 139 from both sides.; minus 125 cos ⁡ open parenthesis θ close parenthesis, equals 101, Divide both sides by minus 125.; cos ⁡ open parenthesis θ close parenthesis, equals minus 0.808")[$139 − 125 cos ( θ ) & = 240 &  & "Subtract 139 from both sides." \ − 125 cos ( θ ) & = 101 &  & "Divide both sides by " − 125. \ cos ( θ ) & = − 0.808$]
      We have solved equations like this one before:  we use the inverse cosine to solve for #math.equation(block: false, alt: "θ")[$θ$].  Remember that there are two angles between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "360 degrees")[$360^(∘)$] that have a cosine of #math.equation(block: false, alt: "minus 0.808")[$− 0.808$], one in the second quadrant and one in the third quadrant.  The calculator will give us only the second quadrant solution.


      #math.equation(block: true, alt: "θ equals cos to the power minus 1 ⁡ open parenthesis minus 0.808 close parenthesis equals 143.9 degrees")[$θ = cos^(− 1) ( − 0.808 ) = 143.9^(∘)$]
      To find the second solution, we need the third-quadrant angle whose cosine is #math.equation(block: false, alt: "minus 0.808")[$− 0.808$].


      Now, the reference angle for #math.equation(block: false, alt: "143.9 degrees")[$143.9^(∘)$] is


      #math.equation(block: true, alt: "180 degrees minus 143.9 degrees equals 36.1 degrees")[$180^(∘) − 143.9^(∘) = 36.1^(∘)$]
      and the angle in the third quadrant with the same reference angle is


      #math.equation(block: true, alt: "180 degrees plus 36.1 degrees equals 216.1 degrees")[$180^(∘) + 36.1^(∘) = 216.1^(∘)$]
      (See the figure at right.)  Thus, the other solution is #math.equation(block: false, alt: "216.1 degrees")[$216.1^(∘)$].


      #figure(figph[angles], alt: "angles", caption: none)


      To solve simple equations involving a single trigonometric ratio (either #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis , cos ⁡ open parenthesis θ close parenthesis ,")[$sin ( θ ) , cos ( θ ) ,$] or #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$]), we can follow the steps below.


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[To Solve a Trigonometric Equation for #math.equation(block: false, alt: "0 degrees less than or equal to θ less than or equal to 360 degrees")[$0^(∘) ≤ θ ≤ 360^(∘)$]]


        + Isolate the trigonometric ratio.
+ Find one solution.+ Give an exact solution if the trig ratio is one of the special values.
+ Otherwise use the inverse trig keys on a calculator.
+ Use reference angles to find a second solution (if there is one).
]


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
In step 3, note that if the first solution is a quadrantal angle, there may not be a second solution. Consider, for example, the equation #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals minus 1")[$cos ( θ ) = − 1$].
]


      To finish up our Ferris wheel example, we can replace #math.equation(block: false, alt: "θ")[$θ$] by #math.equation(block: false, alt: "the fraction 2 t over 3")[$display(frac(2 t, 3))$] to get two equations:


      #math.equation(block: true, alt: "the fraction 2 t over 3, equals 143.9; 2 t, equals 431.7; t, equals 215.35")[$display(frac(2 t, 3)) & = 143.9 \ 2 t & = 431.7 \ t & = 215.35$]
      #math.equation(block: true, alt: "the fraction 2 t over 3, equals 216.1; 2 t, equals 648.3; t, equals 324.5")[$display(frac(2 t, 3)) & = 216.1 \ 2 t & = 648.3 \ t & = 324.5$]
      Your height is above 240 feet between 215.35 seconds and 324.15 seconds. Our estimates from the graphical solutions were pretty good.


      #examplebox("Example 3")[][
Solve the equation #math.equation(block: false, alt: "8 sin ⁡ open parenthesis θ close parenthesis minus 1 equals 3")[$" " " " 8 sin ( θ ) − 1 = 3 " " " "$] for #math.equation(block: false, alt: "0 degrees less than or equal to θ less than or equal to 360 degrees")[$0^(∘) ≤ θ ≤ 360^(∘)$].


        
        #solutionbox[
We begin by isolating the trig ratio.


          #math.equation(block: true, alt: "8 sin ⁡ open parenthesis θ close parenthesis minus 1, equals 3, Add 1 to both sides.; 8 sin ⁡ open parenthesis θ close parenthesis, equals 4, Divide both sides by 8.; sin ⁡ open parenthesis θ close parenthesis, equals the fraction 4 over 8 equals the fraction 1 over 2")[$8 sin ( θ ) − 1 & = 3 &  & "Add 1 to both sides." \ 8 sin ( θ ) & = 4 &  & "Divide both sides by 8." \ sin ( θ ) & = display(frac(4, 8)) = display(frac(1, 2))$]
          We recognize that #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 1 over 2")[$sin ( θ ) = display(frac(1, 2))$] for one of the special angles, namely #math.equation(block: false, alt: "θ equals 30 degrees")[$θ = 30^(∘)$], and also for the obtuse angle #math.equation(block: false, alt: "θ equals 150 degrees")[$θ = 150^(∘)$], as shown at right.


          #figure(figph[special angles], alt: "special angles", caption: none)


          We know that #math.equation(block: false, alt: "θ equals 150 degrees")[$θ = 150^(∘)$] is the second-quadrant angle with reference angle#math.equation(block: false, alt: "θ equals 30 degrees")[$θ = 30^(∘)$]. What about the other quadrants? Because the sine is negative in the third and fourth quadrants, there are no other solutions between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "360 degrees")[$360^(∘)$].  The solutions we want are #math.equation(block: false, alt: "θ equals 30 degrees")[$θ = 30^(∘)$] and #math.equation(block: false, alt: "θ equals 150 degrees")[$θ = 150^(∘)$].
]
]


      #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Be careful! It is worth repeating that to solve a trigonometric equation, it is not enough to find one solution; we must find #emph[all] the solutions between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "360 degrees")[$360^(∘)$]. To find the solutions in other quadrants, we use reference angles.
]


      In the previous Example, we found two solutions of the equation #math.equation(block: false, alt: "8 sin ⁡ open parenthesis θ close parenthesis minus 1 equals 3")[$8 sin ( θ ) − 1 = 3$].  The equation actually has infinitely many solutions, as you can see in the figure below, which shows a graph of #math.equation(block: false, alt: "y equals 8 sin ⁡ open parenthesis θ close parenthesis minus 1")[$y = 8 sin ( θ ) − 1$]  and the horizontal line #math.equation(block: false, alt: "y equals 3")[$y = 3$].  The line intersects the sine graph infinitely many times, twice in each cycle.


      #figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none)


      Each intersection represents a solution of the equation. Thus, all of the angles coterminal with #math.equation(block: false, alt: "30 degrees")[$30^(∘)$] and #math.equation(block: false, alt: "150 degrees")[$150^(∘)$] are also solutions. We can easily find these solutions by adding integer multiples of #math.equation(block: false, alt: "360 degrees")[$360^(∘)$] to #math.equation(block: false, alt: "30 degrees")[$30^(∘)$] or #math.equation(block: false, alt: "150 degrees")[$150^(∘)$]. This is why, when solving a trigonometric equation, we usually list only the solutions in one cycle, typically those between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "360 degrees")[$360^(∘)$].


      
        Solve the equation #math.equation(block: false, alt: "5 tan ⁡ open parenthesis θ close parenthesis plus 7 equals 2")[$" " " " 5 tan ( θ ) + 7 = 2 " " " "$] for #math.equation(block: false, alt: "0 degrees less than or equal to θ less than or equal to 360 degrees")[$0^(∘) ≤ θ ≤ 360^(∘)$].


        
          We first isolate the trig ratio to get  #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals minus 1")[$tan ( θ ) = − 1$]. We recognize the special angle #math.equation(block: false, alt: "tan ⁡ open parenthesis 45 degrees close parenthesis equals 1")[$tan ( 45^(∘) ) = 1$].  Because the tangent is negative in the second and fourth quadrants, we find the solutions #math.equation(block: false, alt: "θ equals 135 degrees and θ equals 315 degrees")[$θ = 135^(∘) " " "and" " " θ = 315^(∘)$].


        
      
      We can use a calculator to help us solve equations that do not involve special angles.


      #examplebox("Example 4")[][
Solve the equation #math.equation(block: false, alt: "3 tan ⁡ open parenthesis β close parenthesis plus 1 equals minus 8")[$" " " " 3 tan ( β ) + 1 = − 8 " " " "$] for #math.equation(block: false, alt: "0 degrees less than or equal to θ less than or equal to 360 degrees")[$0^(∘) ≤ θ ≤ 360^(∘)$]. Round your solutions to three decimal places.


        
        #solutionbox[
First, we isolate the trig ratio.


          #math.equation(block: true, alt: "3 tan ⁡ open parenthesis β close parenthesis plus 1, equals minus 8; 3 tan ⁡ open parenthesis β close parenthesis, equals minus 9; tan ⁡ open parenthesis β close parenthesis, equals minus 3")[$3 tan ( β ) + 1 & = − 8 \ 3 tan ( β ) & = − 9 \ tan ( β ) & = − 3$]
          There are two angles with tangent #math.equation(block: false, alt: "minus 3")[$− 3$], one in the second quadrant and one in the fourth quadrant.  The calculator finds the solution


          #math.equation(block: true, alt: "tan to the power minus 1 ⁡ open parenthesis minus 3 close parenthesis equals minus 71.565 degrees")[$tan^(− 1) ( − 3 ) = − 71.565^(∘)$]
          This angle is in the fourth quadrant, but it is not between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "360 degrees")[$360^(∘)$].


          #figure(figph[angles], alt: "angles", caption: none)


          The angle we want is coterminal with #math.equation(block: false, alt: "minus 71.565 degrees")[$− 71.565^(∘)$], so we add #math.equation(block: false, alt: "360 degrees")[$360^(∘)$] to get the first solution:


          #math.equation(block: true, alt: "B equals tan to the power minus 1 ⁡ open parenthesis minus 3 close parenthesis plus 360 degrees equals 288.435 degrees")[$B = tan^(− 1) ( − 3 ) + 360^(∘) = 288.435^(∘)$]
          The other solution is in the second quadrant, as shown in the figure.  This angle is #math.equation(block: false, alt: "180 degrees")[$180^(∘)$] less than the fourth quadrant solution:


          #math.equation(block: true, alt: "B equals tan to the power minus 1 ⁡ open parenthesis minus 3 close parenthesis plus 180 degrees equals 108.435 degrees")[$B = tan^(− 1) ( − 3 ) + 180^(∘) = 108.435^(∘)$]
]
]


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
In the previous Example, notice that the solutions are #math.equation(block: false, alt: "180 degrees")[$180^(∘)$] apart. The solutions of an equation #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals k")[$tan ( θ ) = k$] always differ by #math.equation(block: false, alt: "180 degrees")[$180^(∘)$], so the use of reference angles is not necessary for these equations:  once we have found one solution, we can add or subtract #math.equation(block: false, alt: "180 degrees")[$180^(∘)$] to find the other solution between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "360 degrees")[$360^(∘)$].  (See Homework Problems 69 and 70.)
]


      
        Solve the equation #math.equation(block: false, alt: "4 cos ⁡ open parenthesis C close parenthesis minus 6 equals minus 3")[$" " " " 4 cos ( C ) − 6 = − 3$], for #math.equation(block: false, alt: "0 degrees less than or equal to C less than or equal to 360 degrees")[$0^(∘) ≤ C ≤ 360^(∘)$]. Round your solutions to three decimal places.


        
          We isolate the trig ratio to find #math.equation(block: false, alt: "cos ⁡ open parenthesis C close parenthesis equals the fraction 3 over 4")[$cos ( C ) = frac(3, 4)$], and #math.equation(block: false, alt: "cos to the power minus 1 ⁡ open parenthesis the fraction 3 over 4 close parenthesis equals 41.4 degrees")[$cos^(− 1) ( frac(3, 4) ) = 41.4^(∘)$]. The second solution lies in the fourthe quadrant at #math.equation(block: false, alt: "360 degrees minus 41.4 degrees equals 318.6 degrees")[$360^(∘) − 41.4^(∘) = 318.6^(∘)$].


        
      
      Some trigonometric equations have no solution. As we can observe from their graphs or from their definitions, the sine and cosine functions only have values ranging from #math.equation(block: false, alt: "minus 1")[$− 1$] to #math.equation(block: false, alt: "1")[$1$].


      #examplebox("Example 5")[][
Solve  #math.equation(block: false, alt: "the fraction sin ⁡ open parenthesis β close parenthesis over 5 minus 3 equals 1")[$" " " " display(frac(sin ( β ), 5)) − 3 = 1 " " " "$] for #math.equation(block: false, alt: "0 degrees less than or equal to θ less than or equal to 360 degrees")[$0^(∘) ≤ θ ≤ 360^(∘)$].


        
        #solutionbox[
We begin by isolating #math.equation(block: false, alt: "sin ⁡ open parenthesis β close parenthesis")[$sin ( β )$].


          #math.equation(block: true, alt: "the fraction sin ⁡ open parenthesis β close parenthesis over 5 minus 3, equals 1, Add 3 to both sides.; the fraction sin ⁡ open parenthesis β close parenthesis over 5, equals 4, Multiply both sides by 5.; sin ⁡ open parenthesis β close parenthesis, equals 20")[$display(frac(sin ( β ), 5)) − 3 & = 1 &  & "Add 3 to both sides." \ display(frac(sin ( β ), 5)) & = 4 &  & "Multiply both sides by 5." \ sin ( β ) & = 20$]
          Because #math.equation(block: false, alt: "sin ⁡ open parenthesis β close parenthesis")[$sin ( β )$]  is never greater than 1, there is no angle #math.equation(block: false, alt: "β")[$β$] whose sine is 20. The equation has no solution.
]
]


      
        Solve #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis plus 5 equals 3")[$" " " " cos ( θ ) + 5 = 3 " " " "$] for #math.equation(block: false, alt: "0 degrees less than or equal to θ less than or equal to 360 degrees")[$0^(∘) ≤ θ ≤ 360^(∘)$].


        
          Because #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$" " " " cos ( θ )$] cannot equal #math.equation(block: false, alt: "minus 2")[$− 2$], there is no solution.


        
      
    
    
      
=== Graphical Solutions


      Sometimes it is helpful to have a visual image of an equation, and we can use graphs to find approximate solutions.


      #examplebox("Example 6")[][
Use a graph to solve the equation #math.equation(block: false, alt: "3 tan ⁡ open parenthesis B close parenthesis plus 1 equals minus 8")[$" " " " 3 tan ( B ) + 1 = − 8$].


        
        #solutionbox[
Graph the equation #math.equation(block: false, alt: "y equals 3 tan ⁡ open parenthesis B close parenthesis plus 1")[$" " " " y = 3 tan ( B ) + 1 " " " "$] and the horizontal line #math.equation(block: false, alt: "y equals minus 8")[$y = − 8$] on the same axes, as shown below.


          #figure(figph[tangent graph], alt: "tangent graph", caption: none)


          It is difficult to read the graph with much accuracy, but you can use the intersect feature on your calculator to verify that the graph of #math.equation(block: false, alt: "y equals 3 tan ⁡ open parenthesis B close parenthesis plus 1")[$" " " " y = 3 tan ( B ) + 1 " " " "$] has #math.equation(block: false, alt: "y")[$y$]-value #math.equation(block: false, alt: "minus 8")[$− 8$] when #math.equation(block: false, alt: "B")[$B$] is about #math.equation(block: false, alt: "108.4 degrees")[$108.4^(∘)$] or #math.equation(block: false, alt: "288.4 degrees")[$288.4^(∘)$].
]
]


      
        Use a graph to verify your solutions to #math.equation(block: false, alt: "4 cos ⁡ open parenthesis C close parenthesis minus 6 equals minus 3")[$" " " " 4 cos ( C ) − 6 = − 3$] from Exercise.


        
          #figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none)


          The graphs intersect at about #math.equation(block: false, alt: "40 degrees")[$40^(∘)$] and about #math.equation(block: false, alt: "320 degrees")[$320^(∘)$].


        
      
    
    
      
=== Equations with Squares of Trig Ratios


      You know several techniques for solving quadratic equations. Simple quadratic equations can be solved by extracting roots.  For example, to solve the equation


      #math.equation(block: true, alt: "4 x squared plus 3 equals 15")[$4 x^(2) + 3 = 15$]
      we first isolate #math.equation(block: false, alt: "x squared")[$x^(2)$]:


      #math.equation(block: true, alt: "4 x squared, equals 12; x squared, equals 3")[$4 x^(2) & = 12 \ x^(2) & = 3$]
      and then take square roots of both sides to find


      #math.equation(block: true, alt: "x equals plus or minus the square root of 3 approximately equals plus or minus 1.732")[$x = ± sqrt(3) ≈ ± 1.732$]
      Recall that a quadratic equation may have two real solutions, one (repeated) real solution, or no real solutions.  We can use extraction of roots to solve trigonometric equations as well.


      #examplebox("Example 7")[][
Solve #math.equation(block: false, alt: "4 tan squared ⁡ open parenthesis θ close parenthesis plus 3 equals 15")[$" " " " 4 tan^(2) ( θ ) + 3 = 15 " " " "$] for #math.equation(block: false, alt: "0 degrees less than or equal to θ less than or equal to 360 degrees")[$0^(∘) ≤ θ ≤ 360^(∘)$].


        
        #solutionbox[
We begin by isolating the power of tangent, #math.equation(block: false, alt: "tan squared ⁡ θ")[$tan^(2) θ$].


          #math.equation(block: true, alt: "4 tan squared ⁡ open parenthesis θ close parenthesis, equals 12; tan squared ⁡ open parenthesis θ close parenthesis, equals 3")[$4 tan^(2) ( θ ) & = 12 \ tan^(2) ( θ ) & = 3$]
          Next, we solve for #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$]  by extracting roots.


          #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals plus or minus the square root of 3")[$tan ( θ ) = ± sqrt(3)$]
          There are two angles between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "360 degrees")[$360^(∘)$] with tangent #math.equation(block: false, alt: "the square root of 3")[$sqrt(3)$] and two angles with tangent #math.equation(block: false, alt: "minus the square root of 3")[$− sqrt(3)$], making four solutions to this equation. We know that #math.equation(block: false, alt: "tan ⁡ open parenthesis 60 degrees close parenthesis equals the square root of 3")[$tan ( 60^(∘) ) = sqrt(3)$],  so one of the solutions is #math.equation(block: false, alt: "60 degrees")[$60^(∘)$].


          The tangent function is also positive in the third quadrant, and the angle in the third quadrant with reference angle #math.equation(block: false, alt: "60 degrees")[$60^(∘)$] is #math.equation(block: false, alt: "180 degrees plus 60 degrees equals 240 degrees")[$180^(∘) + 60^(∘) = 240^(∘)$]. The angles in the second and fourth quadrants with reference angle #math.equation(block: false, alt: "60 degrees")[$60^(∘)$] have tangent #math.equation(block: false, alt: "minus the square root of 3")[$− sqrt(3)$].  These angles are


          #figure(figph[angles], alt: "angles", caption: none)


          #math.equation(block: true, alt: "180 degrees minus 60 degrees equals 120 degrees and 360 degrees minus 60 degrees equals 300 degrees")[$180^(∘) − 60^(∘) = 120^(∘) " " " " "and" " " " " 360^(∘) − 60^(∘) = 300^(∘)$]
          The four solutions are shown at right.
]
]


      
        Solve the equation #math.equation(block: false, alt: "2 sin squared ⁡ open parenthesis θ close parenthesis equals 1")[$" " " " 2 sin^(2) ( θ ) = 1 " " " "$] for #math.equation(block: false, alt: "0 degrees less than or equal to θ less than or equal to 360 degrees")[$0^(∘) ≤ θ ≤ 360^(∘)$].


        
          #math.equation(block: false, alt: "θ equals 45 degrees")[$θ = 45^(∘)$], #math.equation(block: false, alt: "θ equals 135 degrees")[$" " θ = 135^(∘)$], #math.equation(block: false, alt: "θ equals 225 degrees")[$" " θ = 225^(∘) " "$],  or #math.equation(block: false, alt: "θ equals 315 degrees")[$" " θ = 315^(∘)$]


        
      
      Other quadratic equations can be solved by factoring. For example, we can solve the equation


      #math.equation(block: true, alt: "4 x squared plus 4 x minus 3 equals 0")[$4 x^(2) + 4 x − 3 = 0$]
      by factoring the left side to get


      #math.equation(block: true, alt: "open parenthesis 2 x plus 3 close parenthesis open parenthesis 2 x minus 1 close parenthesis equals 0")[$( 2 x + 3 ) ( 2 x − 1 ) = 0$]
      Then we apply the Zero Factor Principle to set each factor equal to zero, and solve each equation.


      #math.equation(block: true, alt: "2 x plus 3, equals 12, 2 x minus 1, equals 0; x, equals the fraction minus 3 over 2, x, equals the fraction 1 over 2")[$2 x + 3 & = 12 & 2 x − 1 & = 0 \ x & = display(frac(− 3, 2)) & x & = display(frac(1, 2))$]
      The solutions are #math.equation(block: false, alt: "the fraction minus 3 over 2")[$display(frac(− 3, 2))$] and #math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$].


      #examplebox("Example 8")[][
Solve #math.equation(block: false, alt: "2 cos squared ⁡ open parenthesis θ close parenthesis minus cos ⁡ open parenthesis θ close parenthesis minus 1 equals 0")[$" " " " 2 cos^(2) ( θ ) − cos ( θ ) − 1 = 0 " " " "$] for #math.equation(block: false, alt: "0 degrees less than or equal to θ less than or equal to 360 degrees")[$0^(∘) ≤ θ ≤ 360^(∘)$].


        
        #solutionbox[
Compare this equation to the algebraic equation #math.equation(block: false, alt: "2 x squared minus x minus 1 equals 0")[$" " 2 x^(2) − x − 1 = 0$]. Our equation has the same algebraic form, but with #math.equation(block: false, alt: "x")[$x$] replaced by #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$].


          We can solve the algebraic equation by factoring #math.equation(block: false, alt: "2 x squared minus x minus 1")[$" " 2 x^(2) − x − 1 " "$] as #math.equation(block: false, alt: "open parenthesis 2 x plus 1 close parenthesis open parenthesis x minus 1 close parenthesis ,")[$" " ( 2 x + 1 ) ( x − 1 ) ,$] and we'll use the same strategy on the trigonometric equation.


          #math.equation(block: true, alt: "2 cos squared ⁡ open parenthesis θ close parenthesis minus cos ⁡ open parenthesis θ close parenthesis minus 1, equals 0, Factor the left side.; open parenthesis 2 cos ⁡ open parenthesis θ close parenthesis plus 1 close parenthesis open parenthesis cos ⁡ open parenthesis θ close parenthesis minus 1 close parenthesis, equals 0, Set each factor equal to zero.; cos ⁡ open parenthesis θ close parenthesis equals the fraction minus 1 over 2 or cos ⁡ open parenthesis θ close parenthesis, equals 1")[$2 cos^(2) ( θ ) − cos ( θ ) − 1 & = 0 &  & "Factor the left side." \ ( 2 cos ( θ ) + 1 ) ( cos ( θ ) − 1 ) & = 0 &  & "Set each factor equal to zero." \ cos ( θ ) = display(frac(− 1, 2)) " " " " " or " " " " " cos ( θ ) & = 1 &$]
          Now we solve each equation for #math.equation(block: false, alt: "θ")[$θ$]. We know that #math.equation(block: false, alt: "cos ⁡ open parenthesis 60 degrees close parenthesis equals the fraction 1 over 2 ,")[$cos ( 60^(∘) ) = display(frac(1, 2)) ,$] and the cosine function is negative in the second and third quadrants. The angles in those quadrants with reference angle #math.equation(block: false, alt: "60 degrees")[$60^(∘)$] are


          #math.equation(block: true, alt: "180 degrees minus 60 degrees equals 120 degrees and 180 degrees plus 60 degrees equals 240 degrees")[$180^(∘) − 60^(∘) = 120^(∘) " " " " " " "and" " " " " " " 180^(∘) + 60^(∘) = 240^(∘)$]
          Also, #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals 1")[$cos ( θ ) = 1$]  when #math.equation(block: false, alt: "θ equals 0 degrees")[$θ = 0^(∘)$]. Thus, the original equation has three solutions: #math.equation(block: false, alt: "θ equals 120 degrees , θ equals 240 degrees ,")[$θ = 120^(∘) , " " θ = 240^(∘) ,$] and #math.equation(block: false, alt: "θ equals 0 degrees .")[$θ = 0^(∘) .$]
]
]


      
        Solve #math.equation(block: false, alt: "9 sin squared ⁡ open parenthesis θ close parenthesis minus 3 sin ⁡ open parenthesis θ close parenthesis equals 2")[$" " " " 9 sin^(2) ( θ ) − 3 sin ( θ ) = 2 " " " "$] for #math.equation(block: false, alt: "0 degrees less than or equal to θ less than or equal to 360 degrees")[$0^(∘) ≤ θ ≤ 360^(∘)$]. Round your answers to the nearest degree.


        
          We first solve the equation #math.equation(block: false, alt: "9 sin squared ⁡ open parenthesis θ close parenthesis minus 3 sin ⁡ open parenthesis θ close parenthesis minus 2 equals 0")[$" " 9 sin^(2) ( θ ) − 3 sin ( θ ) − 2 = 0 " "$] by factoring:


          #math.equation(block: true, alt: "open parenthesis 3 sin ⁡ open parenthesis θ close parenthesis minus 2 close parenthesis open parenthesis 3 sin ⁡ open parenthesis θ close parenthesis plus 1 close parenthesis equals 0")[$( 3 sin ( θ ) − 2 ) ( 3 sin ( θ ) + 1 ) = 0$]
          , 
    so #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 2 over 3")[$sin ( θ ) = display(frac(2, 3))$] or #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction minus 1 over 3")[$sin ( θ ) = display(frac(− 1, 3))$]. From the first equation we find #math.equation(block: false, alt: "θ equals 42 degrees")[$θ = 42^(∘)$] and #math.equation(block: false, alt: "θ equals 138 degrees")[$" " θ = 138^(∘)$], and from the second euqtion we find #math.equation(block: false, alt: "θ equals 199 degrees")[$θ = 199^(∘) " "$] and #math.equation(block: false, alt: "θ equals 341 degrees")[$" " θ = 341^(∘)$].


        
      
    
    
      
=== Snell's Law


      When you view an object through a liquid, such as a spoon in a glass of water, or a fish in an aquarium, the object may look distorted or bent. This distortion is caused by refraction of light. Light rays bend when they pass from one medium to another, for instance from water to glass or from glass to air.


      #figure(figph[spoon in glass of water], alt: "spoon in glass of water", caption: none)


      A light ray enters the boundary between the two media at a certain angle, called the #strong[angle of incidence], but leaves the boundary at a different angle, the #strong[angle of refraction].  Both angles are acute angles measured from the #strong[normal] line perpendicular to the boundary, as shown below.


      The change of angle is caused by the fact that light travels at different speeds in different media.  The relationship between the angle of incidence and the angle of refraction is given by Snell's Law:


      #math.equation(block: true, alt: "the fraction sin ⁡ open parenthesis θ sub 1 close parenthesis over sin ⁡ open parenthesis θ sub 2 close parenthesis equals the fraction v sub 1 over v sub 2")[$display(frac(bold(sin) ( bold(italic(θ))_(1) ), bold(sin) ( bold(italic(θ))_(2) ))) = display(frac(bold(italic(v))_(1), bold(italic(v))_(2)))$]
      #figure(figph[water-glass interface], alt: "water-glass interface", caption: none)


      where #math.equation(block: false, alt: "θ sub 1")[$θ_(1)$] is the angle in the medium where light travels at speed #math.equation(block: false, alt: "v sub 1")[$v_(1)$], and #math.equation(block: false, alt: "θ sub 2")[$θ_(2)$] is the angle where light travels at speed  #math.equation(block: false, alt: "v sub 2")[$v_(2)$]. The ratio of the speeds is called the #strong[index of refraction].


      #examplebox("Example 9")[][
The index of refraction from water to glass is 1.1.  If light passes from water to glass with a #math.equation(block: false, alt: "23 degrees")[$23^(∘)$] angle of incidence, what is the angle of refraction?


        
        #solutionbox[
If #math.equation(block: false, alt: "θ")[$θ$] is the angle of refraction, then from Snell's Law we have


          #math.equation(block: true, alt: "the fraction sin ⁡ open parenthesis 23 degrees close parenthesis over sin ⁡ open parenthesis θ close parenthesis, equals 1.1, Multiply both sides by sin ⁡ open parenthesis θ close parenthesis .; sin ⁡ open parenthesis 23 degrees close parenthesis, equals 1.1 sin ⁡ open parenthesis θ close parenthesis, Divide both sides by 1.1.; sin ⁡ open parenthesis θ close parenthesis, equals the fraction sin ⁡ open parenthesis 23 degrees close parenthesis over 1.1 equals 0.3552")[$display(frac(sin ( 23^(∘) ), sin ( θ ))) & = 1.1 &  & "Multiply both sides by" bold(sin) ( bold(italic(θ)) ) . \ sin ( 23^(∘) ) & = 1.1 sin ( θ ) &  & "Divide both sides by 1.1." \ sin ( θ ) & = display(frac(sin ( 23^(∘) ), 1.1)) = 0.3552$]
          Because #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals 0.3552")[$sin ( θ ) = 0.3552$], #math.equation(block: false, alt: "θ equals sin to the power minus 1 ⁡ open parenthesis 0.3552 close parenthesis equals 20.8 degrees .")[$" " θ = sin^(− 1) ( 0.3552 ) = 20.8^(∘) .$]  (For Snell's Law we use only acute angles.) The angle of refraction is approximately #math.equation(block: false, alt: "20.8 degrees")[$20.8^(∘)$].
]
]


      
        A light ray passes from water to glass with an #math.equation(block: false, alt: "18 degrees")[$18^(∘)$] angle of incidence.  What is the angle of refraction?


        
          #math.equation(block: true, alt: "16.315 degrees")[$16.315^(∘)$]


        
      
      Review the following skills you will need for this section.


      #notebox("Algebra Refresher", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Each of the following "solutions" contains an error.  Find the error, and supply a correct solution.


        + #math.equation(block: false, alt: "3 x squared minus 5 x, equals 0; 3 x squared, equals 5 x; 3 x, equals 5; x, equals the fraction 5 over 3")[$3 x^(2) − 5 x & = 0 \ 3 x^(2) & = 5 x \ 3 x & = 5 \ x & = display(frac(5, 3))$]
+ #math.equation(block: false, alt: "4 x squared, equals 12; x squared, equals 3; x, equals the square root of 3")[$4 x^(2) & = 12 \ x^(2) & = 3 \ x & = sqrt(3)$]
+ #math.equation(block: false, alt: "open parenthesis 2 x plus 1 close parenthesis squared, equals 4; 2 x plus 1, equals 2; 2 x, equals 1; x, equals the fraction 1 over 2")[$( 2 x + 1 )^(2) & = 4 \ 2 x + 1 & = 2 \ 2 x & = 1 \ x & = display(frac(1, 2))$]
+ #math.equation(block: false, alt: "minus 2 x squared minus 4 x, equals 3; minus 2 x open parenthesis x plus 2 close parenthesis, equals 3; minus 2 x equals 3, x plus 2 equals 3; x equals the fraction minus 3 over 2, x equals 1")[$− 2 x^(2) − 4 x & = 3 \ − 2 x ( x + 2 ) & = 3 \ − 2 x = 3 #h(2em) & x + 2 = 3 \ x = display(frac(− 3, 2)) #h(2em) & x = 1$]


        #math.equation(block: true, alt: "bar")[$underline(#h(2em) #h(2em) #h(2em) #h(2em))$]


        Algebra Refresher Answers


        + We can't divide by #math.equation(block: false, alt: "x . x equals 0 , the fraction 5 over 3")[$x . " " " " x = 0 , display(frac(5, 3))$]
+ 3 has two square roots. #math.equation(block: false, alt: "x equals plus or minus the square root of 3")[$" " x = ± sqrt(3)$]
+ 4 has two square roots. #math.equation(block: false, alt: "x equals the fraction minus 3 over 2 , the fraction 1 over 2")[$" " x = display(frac(− 3, 2)) , display(frac(1, 2))$]
+ One side must be zero to use the zero-factor principle.  No real solutions.
]


    
    
      
=== Section 5.2 Summary


      
        
==== Vocabulary


        - Equation
- Solve
- Zero Factor Principle
- Angle of incidence
- Angle of refraction
- Normal


      
      
        
==== Concepts


        + An #strong[equation] is a statement that two algebraic expressions are equal.  It may be true or false.
+ We can solve equations by trial and error, by using graphs, or by algebraic techniques.
+ To solve a trigonometric equation, we first isolate the trigonometric ratio on one side of the equation.
+ We use reference angles to find all the solutions between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "360 degrees")[$360^(∘)$].
+ We can use factoring or extraction of roots to solve some quadratic equations.


      
      
        
====  Study Questions


        + How many solutions between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "360 degrees")[$360^(∘)$] does the equation #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals k")[$cos ( θ ) = k$]  have for each value of #math.equation(block: false, alt: "k")[$k$] between -1 and 0?
+ How many solutions between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "360 degrees")[$360^(∘)$] does the equation #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals k")[$cos ( θ ) = k$]  have for each value of #math.equation(block: false, alt: "k")[$k$] greater than 1?
+ How many solutions between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "360 degrees")[$360^(∘)$] does the equation #math.equation(block: false, alt: "sin squared ⁡ open parenthesis θ close parenthesis equals k")[$sin^(2) ( θ ) = k$] have for each value of #math.equation(block: false, alt: "k")[$k$] between -1 and 0?


      
      
        
==== Skills


        + Use reference angles    \#1–8
+ Solve equations by trial and error    \#9–14
+ Use graphs to solve equations    \#15–18, \#39–52
+ Solve trigonometric equations for exact values    \#19–32, 39–46
+ Use a calculator to solve trigonometric equations    \#33–38, 47–52, 65–68
+ Solve trigonometric equations that involve factoring    \#53–64


      
    
    
      
=== Homework 5.2


      For Problems 1–4, find the reference angle. (If you would like to review reference angles, see Section 4.1.)


      
        #math.equation(block: true, alt: "250 degrees")[$250^(∘)$]


        
          #math.equation(block: true, alt: "70 degrees")[$70^(∘)$]


        
      
      
        #math.equation(block: true, alt: "145 degrees")[$145^(∘)$]


      
      
        #math.equation(block: true, alt: "320 degrees")[$320^(∘)$]


        
          #math.equation(block: true, alt: "40 degrees")[$40^(∘)$]


        
      
      
        #math.equation(block: true, alt: "minus 110 degrees")[$− 110^(∘)$]


      
      For Problems 5–8, find an angle in each quadrant with the given reference angle.


      
        #math.equation(block: true, alt: "18 degrees")[$18^(∘)$]


        
          I: #math.equation(block: false, alt: "18 degrees ;")[$18^(∘) ;$]  II: #math.equation(block: false, alt: "162 degrees ;")[$162^(∘) ;$]  III: #math.equation(block: false, alt: "198 degrees ;")[$198^(∘) ;$]  IV: #math.equation(block: false, alt: "342 degrees")[$342^(∘)$]


        
      
      
        #math.equation(block: true, alt: "35 degrees")[$35^(∘)$]


      
      
        #math.equation(block: true, alt: "52 degrees")[$52^(∘)$]


        
          I: #math.equation(block: false, alt: "52 degrees ;")[$52^(∘) ;$]  II: #math.equation(block: false, alt: "128 degrees ;")[$128^(∘) ;$]  III: #math.equation(block: false, alt: "232 degrees ;")[$232^(∘) ;$]  IV: #math.equation(block: false, alt: "308 degrees")[$308^(∘)$]


        
      
      
        #math.equation(block: true, alt: "78 degrees")[$78^(∘)$]


      
      For Problems 9–14,


      + Evaluate the expression at the given values of the variable.
+ Give one solution of the equation.


      
        + #math.equation(block: false, alt: "x cubed minus 3 x squared plus 4 ; x equals minus 1 , 0 , 1 , 2 , 3")[$x^(3) − 3 x^(2) + 4 ; " " " " " " x = − 1 , 0 , 1 , 2 , 3$]
+ #math.equation(block: false, alt: "x cubed minus 3 x squared plus 4 equals 0")[$x^(3) − 3 x^(2) + 4 = 0$]


        
          + #math.equation(block: false, alt: "0 , 4 , 2 , 0 , 4")[$0 , " " 4 , " " 2 , " " 0 , " " 4$]
+ #math.equation(block: false, alt: "minus 1 or 2")[$− 1 " " "or" " " 2$]


        
      
      
        + #math.equation(block: false, alt: "the square root of x plus the square root of 2 x plus 1 ; x equals 0 , 2 , 4 , 6")[$sqrt(x) + sqrt(2 x + 1) ; " " " " " " x = 0 , 2 , 4 , 6$]
+ #math.equation(block: false, alt: "the square root of x plus the square root of 2 x plus 1 equals 5")[$sqrt(x) + sqrt(2 x + 1) = 5$]


      
      
        + #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis plus cos ⁡ open parenthesis θ close parenthesis ; θ equals 0 degrees , 30 degrees , 45 degrees , 60 degrees")[$sin ( θ ) + cos ( θ ) ; " " " " " " θ = 0^(∘) , 30^(∘) , 45^(∘) , 60^(∘)$]
+ #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis plus cos ⁡ open parenthesis θ close parenthesis equals the square root of 2")[$sin ( θ ) + cos ( θ ) = sqrt(2)$]


        
          + #math.equation(block: false, alt: "1 , the fraction the square root of 3 plus 1 over 2 , the square root of 2 , the fraction the square root of 3 plus 1 over 2")[$1 , " " display(frac(sqrt(3) + 1, 2)) , " " sqrt(2) , " " display(frac(sqrt(3) + 1, 2))$]
+ #math.equation(block: false, alt: "45 degrees")[$45^(∘)$]


        
      
      
        + #math.equation(block: false, alt: "sin squared ⁡ open parenthesis α close parenthesis minus cos ⁡ open parenthesis α close parenthesis ; α equals 45 degrees , 90 degrees , 135 degrees , 180 degrees")[$sin^(2) ( α ) − cos ( α ) ; " " " " " " α = 45^(∘) , 90^(∘) , 135^(∘) , 180^(∘)$]
+ #math.equation(block: false, alt: "sin squared ⁡ open parenthesis α close parenthesis minus cos ⁡ open parenthesis α close parenthesis equals 1")[$sin^(2) ( α ) − cos ( α ) = 1$]


      
      
        + #math.equation(block: false, alt: "sin ⁡ open parenthesis β close parenthesis plus 2 cos squared ⁡ open parenthesis β close parenthesis ; β equals 210 degrees , 225 degrees , 240 degrees , 270 degrees")[$sin ( β ) + 2 cos^(2) ( β ) ; " " " " " " β = 210^(∘) , 225^(∘) , 240^(∘) , 270^(∘)$]
+ #math.equation(block: false, alt: "sin ⁡ open parenthesis β close parenthesis plus 2 cos squared ⁡ open parenthesis β close parenthesis equals minus 1")[$sin ( β ) + 2 cos^(2) ( β ) = − 1$]


        
          + #math.equation(block: false, alt: "0 , the fraction 2 minus the square root of 2 over 2 , the fraction 1 minus the square root of 3 over 2 , minus 1")[$0 , " " display(frac(2 − sqrt(2), 2)) , " " display(frac(1 − sqrt(3), 2)) , " " − 1$]
+ #math.equation(block: false, alt: "270 degrees")[$270^(∘)$]


        
      
      
        + #math.equation(block: false, alt: "3 cos squared ⁡ open parenthesis ϕ close parenthesis minus sin squared ⁡ open parenthesis ϕ close parenthesis ; ϕ equals 270 degrees , 300 degrees , 315 degrees , 330 degrees")[$3 cos^(2) ( ϕ ) − sin^(2) ( ϕ ) ; " " " " " " ϕ = 270^(∘) , 300^(∘) , 315^(∘) , 330^(∘)$]
+ #math.equation(block: false, alt: "3 cos squared ⁡ open parenthesis ϕ close parenthesis minus sin squared ⁡ open parenthesis ϕ close parenthesis equals 2")[$3 cos^(2) ( ϕ ) − sin^(2) ( ϕ ) = 2$]


      
      For Problems 15–18, use a graph to solve the equation.  Check your solution by 
    substitution.


      
        #math.equation(block: true, alt: "the fraction minus 1 over 3 x squared plus the fraction 2 over 3 x plus 5 equals 0")[$display(frac(− 1, 3)) x^(2) + display(frac(2, 3)) x + 5 = 0$]


        
          #math.equation(block: true, alt: "x equals 5 , minus 3")[$x = 5 , " " − 3$]


        
      
      
        #math.equation(block: true, alt: "0.0625 x squared plus 0.5 x equals minus 1")[$0.0625 x^(2) + 0.5 x = − 1$]


      
      
        #math.equation(block: true, alt: "x cubed plus 2 x squared minus 6 equals 2 x squared plus 7 x")[$x^(3) + 2 x^(2) − 6 = 2 x^(2) + 7 x$]


        
          #math.equation(block: true, alt: "x equals minus 3 , 1 , 2")[$x = − 3 , " " 1 , " " 2$]


        
      
      
        #math.equation(block: true, alt: "8 minus 12 x plus 6 x squared minus x cubed")[$8 − 12 x + 6 x^(2) − x^(3)$]


      
      For Problems 19–32, solve the equation exactly for #math.equation(block: false, alt: "0 degrees less than or equal to θ less than 360 degrees")[$0^(∘) ≤ θ < 360^(∘)$].


      
        #math.equation(block: true, alt: "3 tan ⁡ open parenthesis θ close parenthesis equals the square root of 3")[$3 tan ( θ ) = sqrt(3)$]


        
          #math.equation(block: false, alt: "θ equals 30 degrees")[$θ = 30^(∘) " "$] or #math.equation(block: false, alt: "θ equals 210 degrees")[$" " θ = 210^(∘)$]


        
      
      
        #math.equation(block: true, alt: "7 sin ⁡ open parenthesis θ close parenthesis plus 11 equals 11")[$7 sin ( θ ) + 11 = 11$]


      
      
        #math.equation(block: true, alt: "3 equals 5 minus 4 cos ⁡ open parenthesis θ close parenthesis")[$3 = 5 − 4 cos ( θ )$]


        
          #math.equation(block: false, alt: "θ equals 60 degrees")[$θ = 60^(∘) " "$] or #math.equation(block: false, alt: "θ equals 300 degrees")[$" " θ = 300^(∘)$]


        
      
      
        #math.equation(block: true, alt: "6 tan ⁡ open parenthesis θ close parenthesis plus 21 equals 15")[$6 tan ( θ ) + 21 = 15$]


      
      
        #math.equation(block: true, alt: "8 sin ⁡ open parenthesis θ close parenthesis plus 5 equals 1")[$8 sin ( θ ) + 5 = 1$]


        
          #math.equation(block: false, alt: "θ equals 210 degrees")[$θ = 210^(∘) " "$] or #math.equation(block: false, alt: "θ equals 330 degrees")[$" " θ = 330^(∘)$]


        
      
      
        #math.equation(block: true, alt: "9 cos ⁡ open parenthesis θ close parenthesis plus 15 equals 6")[$9 cos ( θ ) + 15 = 6$]


      
      
        #math.equation(block: true, alt: "0 equals the square root of 2 plus 2 sin ⁡ open parenthesis θ close parenthesis")[$0 = sqrt(2) + 2 sin ( θ )$]


        
          #math.equation(block: false, alt: "θ equals 225 degrees")[$θ = 225^(∘) " "$] or #math.equation(block: false, alt: "θ equals 315 degrees")[$" " θ = 315^(∘)$]


        
      
      
        #math.equation(block: true, alt: "the square root of 3 cos ⁡ open parenthesis θ close parenthesis equals minus the fraction 3 over 2")[$sqrt(3) cos ( θ ) = − display(frac(3, 2))$]


      
      
        #math.equation(block: true, alt: "cos squared ⁡ open parenthesis θ close parenthesis minus 1 equals 0")[$cos^(2) ( θ ) − 1 = 0$]


        
          #math.equation(block: false, alt: "θ equals 0 degrees")[$θ = 0^(∘) " "$] or #math.equation(block: false, alt: "θ equals 180 degrees")[$" " θ = 180^(∘)$]


        
      
      
        #math.equation(block: true, alt: "1 minus sin squared ⁡ open parenthesis θ close parenthesis equals 0")[$1 − sin^(2) ( θ ) = 0$]


      
      
        #math.equation(block: true, alt: "4 sin squared ⁡ open parenthesis θ close parenthesis minus 3 equals 0")[$4 sin^(2) ( θ ) − 3 = 0$]


        
          #math.equation(block: false, alt: "θ equals 60 degrees , θ equals 120 degrees , θ equals 240 degrees ,")[$θ = 60^(∘) , " " θ = 120^(∘) , " " θ = 240^(∘) , " "$] or #math.equation(block: false, alt: "θ equals 300 degrees")[$" " θ = 300^(∘)$]


        
      
      
        #math.equation(block: true, alt: "0 equals 1 minus 2 cos squared ⁡ open parenthesis θ close parenthesis")[$0 = 1 − 2 cos^(2) ( θ )$]


      
      
        #math.equation(block: true, alt: "1 minus tan squared ⁡ open parenthesis θ close parenthesis equals 0")[$1 − tan^(2) ( θ ) = 0$]


        
          #math.equation(block: false, alt: "θ equals 45 degrees , θ equals 135 degrees , θ equals 225 degrees ,")[$θ = 45^(∘) , " " θ = 135^(∘) , " " θ = 225^(∘) , " "$] or #math.equation(block: false, alt: "θ equals 315 degrees")[$" " θ = 315^(∘)$]


        
      
      
        #math.equation(block: true, alt: "0 equals 6 tan squared ⁡ open parenthesis θ close parenthesis minus 2")[$0 = 6 tan^(2) ( θ ) − 2$]


      
      For Problems 33–38, solve the equation for #math.equation(block: false, alt: "0 degrees less than or equal to θ less than 360 degrees")[$0^(∘) ≤ θ < 360^(∘)$].  Round your answers to two decimal places.


      
        #math.equation(block: true, alt: "the fraction 1 over 2 tan ⁡ open parenthesis θ close parenthesis minus 1 equals minus 3")[$display(frac(1, 2)) tan ( θ ) − 1 = − 3$]


        
          #math.equation(block: false, alt: "θ equals 104.04 degrees")[$θ = 104.04^(∘) " "$] or #math.equation(block: false, alt: "θ equals 284.04 degrees")[$" " θ = 284.04^(∘)$]


        
      
      
        #math.equation(block: true, alt: "3 tan ⁡ open parenthesis θ close parenthesis minus 2 equals 4")[$3 tan ( θ ) − 2 = 4$]


      
      
        #math.equation(block: true, alt: "3 equals 5 cos ⁡ open parenthesis θ close parenthesis")[$3 = 5 cos ( θ )$]


        
          #math.equation(block: false, alt: "θ equals 53.13 degrees")[$θ = 53.13^(∘) " "$] or #math.equation(block: false, alt: "θ equals 306.87 degrees")[$" " θ = 306.87^(∘)$]


        
      
      
        #math.equation(block: true, alt: "4 equals 6 sin ⁡ open parenthesis θ close parenthesis")[$4 = 6 sin ( θ )$]


      
      
        #math.equation(block: true, alt: "7 sin ⁡ open parenthesis θ close parenthesis plus 2 equals 1")[$7 sin ( θ ) + 2 = 1$]


        
          #math.equation(block: false, alt: "θ equals 188.21 degrees")[$θ = 188.21^(∘) " "$] or #math.equation(block: false, alt: "θ equals 351.79 degrees")[$" " θ = 351.79^(∘)$]


        
      
      
        #math.equation(block: true, alt: "2 equals 5 minus the fraction 1 over 3 tan ⁡ open parenthesis θ close parenthesis")[$2 = 5 − display(frac(1, 3)) tan ( θ )$]


      
      For Problems 39–46,


      + Use a graph to estimate the solutions for angles between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "360 degrees")[$360^(∘)$].
+ Solve the equation algebraically.


      
        #math.equation(block: true, alt: "7 minus tan ⁡ open parenthesis A close parenthesis equals 8")[$7 − tan ( A ) = 8$]


        
          #math.equation(block: false, alt: "A equals 135 degrees")[$A = 135^(∘) " "$] or #math.equation(block: false, alt: "A equals 315 degrees")[$" " A = 315^(∘)$]


        
      
      
        #math.equation(block: true, alt: "6 equals 8 tan ⁡ open parenthesis w close parenthesis minus 2")[$6 = 8 tan ( w ) − 2$]


      
      
        #math.equation(block: true, alt: "5 equals 1 minus 8 sin ⁡ open parenthesis ϕ close parenthesis")[$5 = 1 − 8 sin ( ϕ )$]


        
          #math.equation(block: false, alt: "ϕ equals 210 degrees")[$ϕ = 210^(∘) " "$] or #math.equation(block: false, alt: "ϕ equals 330 degrees")[$" " ϕ = 330^(∘)$]


        
      
      
        #math.equation(block: true, alt: "9 minus 4 sin ⁡ open parenthesis t close parenthesis equals 13")[$9 − 4 sin ( t ) = 13$]


      
      
        #math.equation(block: true, alt: "2 cos ⁡ open parenthesis B close parenthesis minus 2 equals minus 2")[$2 cos ( B ) − 2 = − 2$]


        
          #math.equation(block: true, alt: "B equals 90 degrees or B equals 270 degrees")[$B = 90^(∘) " " "or" " " B = 270^(∘)$]


        
      
      
        #math.equation(block: true, alt: "2 minus 6 cos ⁡ open parenthesis u close parenthesis equals 5")[$2 − 6 cos ( u ) = 5$]


      
      
        #math.equation(block: true, alt: "3 equals 2 sin ⁡ open parenthesis θ close parenthesis plus 4")[$3 = 2 sin ( θ ) + 4$]


        
          #math.equation(block: false, alt: "θ equals 210 degrees")[$θ = 210^(∘) " "$] or #math.equation(block: false, alt: "θ equals 330 degrees")[$" " θ = 330^(∘)$]


        
      
      
        #math.equation(block: true, alt: "5 equals 3 cos ⁡ open parenthesis x close parenthesis plus 5")[$5 = 3 cos ( x ) + 5$]


      
      For Problems 47–52,


      + Use a graph to estimate the solutions for angles between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "360 degrees")[$360^(∘)$].
+ Solve the equation algebraically, rounding angles to the nearest degree.


      
        #math.equation(block: true, alt: "8 sin ⁡ open parenthesis t close parenthesis plus 7 equals 4")[$8 sin ( t ) + 7 = 4$]


        
          #math.equation(block: false, alt: "t equals 202 degrees")[$t = 202^(∘) " "$] or #math.equation(block: false, alt: "t equals 338 degrees")[$" " t = 338^(∘)$]


        
      
      
        #math.equation(block: true, alt: "9 minus 6 cos ⁡ open parenthesis A close parenthesis equals 5")[$9 − 6 cos ( A ) = 5$]


      
      
        #math.equation(block: true, alt: "5 tan ⁡ open parenthesis B close parenthesis minus 4 equals minus 2")[$5 tan ( B ) − 4 = − 2$]


        
          #math.equation(block: true, alt: "B equals 22 degrees or B equals 202 degrees")[$B = 22^(∘) " " "or" " " B = 202^(∘)$]


        
      
      
        #math.equation(block: true, alt: "3 minus 10 tan ⁡ open parenthesis C close parenthesis equals minus 11")[$3 − 10 tan ( C ) = − 11$]


      
      
        #math.equation(block: true, alt: "1 plus 6 cos ⁡ open parenthesis ϕ close parenthesis equals minus 4")[$1 + 6 cos ( ϕ ) = − 4$]


        
          #math.equation(block: false, alt: "ϕ equals 146 degrees")[$ϕ = 146^(∘) " "$] or #math.equation(block: false, alt: "ϕ equals 214 degrees")[$" " ϕ = 214^(∘)$]


        
      
      
        #math.equation(block: true, alt: "4 sin ⁡ open parenthesis u close parenthesis minus 2 equals minus 1")[$4 sin ( u ) − 2 = − 1$]


      
      For Problems 53–64, solve the equation for #math.equation(block: false, alt: "0 degrees less than or equal to θ less than 360 degrees")[$0^(∘) ≤ θ < 360^(∘)$].  Round angles to two decimal places.


      
        #math.equation(block: true, alt: "6 cos squared ⁡ open parenthesis θ close parenthesis equals 2")[$6 cos^(2) ( θ ) = 2$]


        
          #math.equation(block: false, alt: "θ equals 54.74 degrees , θ equals 125.26 degrees , θ equals 234.74 degrees ,")[$θ = 54.74^(∘) , " " θ = 125.26^(∘) , " " θ = 234.74^(∘) , " "$] or #math.equation(block: false, alt: "θ equals 305.26 degrees")[$" " θ = 305.26^(∘)$]


        
      
      
        #math.equation(block: true, alt: "2 minus 7 sin squared ⁡ open parenthesis θ close parenthesis equals 1")[$2 − 7 sin^(2) ( θ ) = 1$]


      
      
        #math.equation(block: true, alt: "5 sin squared ⁡ open parenthesis θ close parenthesis plus sin ⁡ open parenthesis θ close parenthesis equals 0")[$5 sin^(2) ( θ ) + sin ( θ ) = 0$]


        
          #math.equation(block: false, alt: "θ equals 0 degrees")[$θ = 0^(∘)$], #math.equation(block: false, alt: "θ equals 180 degrees")[$" " θ = 180^(∘)$], #math.equation(block: false, alt: "θ equals 191.54 degrees ,")[$" " θ = 191.54^(∘) , " "$] or #math.equation(block: false, alt: "θ equals 348.46 degrees")[$" " θ = 348.46^(∘)$]


        
      
      
        #math.equation(block: true, alt: "4 tan squared ⁡ open parenthesis θ close parenthesis equals tan ⁡ open parenthesis θ close parenthesis")[$4 tan^(2) ( θ ) = tan ( θ )$]


      
      
        #math.equation(block: true, alt: "2 cos squared ⁡ open parenthesis θ close parenthesis plus cos ⁡ open parenthesis θ close parenthesis minus 1 equals 0")[$2 cos^(2) ( θ ) + cos ( θ ) − 1 = 0$]


        
          #math.equation(block: false, alt: "θ equals 60 degrees")[$θ = 60^(∘)$], #math.equation(block: false, alt: "θ equals 180 degrees")[$" " θ = 180^(∘)$], or #math.equation(block: false, alt: "θ equals 300 degrees")[$" " θ = 300^(∘)$]


        
      
      
        #math.equation(block: true, alt: "tan squared ⁡ open parenthesis θ close parenthesis minus 5 tan ⁡ open parenthesis θ close parenthesis plus 6 equals 0")[$tan^(2) ( θ ) − 5 tan ( θ ) + 6 = 0$]


      
      
        #math.equation(block: true, alt: "6 tan squared ⁡ open parenthesis θ close parenthesis minus tan ⁡ open parenthesis θ close parenthesis minus 1 equals 0")[$6 tan^(2) ( θ ) − tan ( θ ) − 1 = 0$]


        
          #math.equation(block: false, alt: "θ equals 26.57 degrees")[$θ = 26.57^(∘)$], #math.equation(block: false, alt: "θ equals 161.57 degrees")[$" " θ = 161.57^(∘)$], #math.equation(block: false, alt: "θ equals 206.57 degrees")[$" " θ = 206.57^(∘)$],  or #math.equation(block: false, alt: "θ equals 341.57 degrees")[$" " θ = 341.57^(∘)$]


        
      
      
        #math.equation(block: true, alt: "10 cos squared ⁡ open parenthesis θ close parenthesis minus 7 cos ⁡ open parenthesis θ close parenthesis plus 1 equals 0")[$10 cos^(2) ( θ ) − 7 cos ( θ ) + 1 = 0$]


      
      
        #math.equation(block: true, alt: "tan squared ⁡ open parenthesis θ close parenthesis minus 2 tan ⁡ open parenthesis θ close parenthesis equals 15")[$tan^(2) ( θ ) − 2 tan ( θ ) = 15$]


        
          #math.equation(block: false, alt: "θ equals 78.69 degrees")[$θ = 78.69^(∘)$], #math.equation(block: false, alt: "θ equals 108.43 degrees")[$" " θ = 108.43^(∘)$], #math.equation(block: false, alt: "θ equals 258.69 degrees")[$" " θ = 258.69^(∘)$], or #math.equation(block: false, alt: "θ equals 288.43 degrees")[$" " θ = 288.43^(∘)$]


        
      
      
        #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals tan squared ⁡ open parenthesis θ close parenthesis minus 20")[$tan ( θ ) = tan^(2) ( θ ) − 20$]


      
      
        #math.equation(block: true, alt: "cos squared ⁡ open parenthesis θ close parenthesis minus 4 cos ⁡ open parenthesis θ close parenthesis plus 3 equals 0")[$cos^(2) ( θ ) − 4 cos ( θ ) + 3 = 0$]


        
          #math.equation(block: true, alt: "θ equals 0 degrees")[$θ = 0^(∘)$]


        
      
      
        #math.equation(block: true, alt: "sin squared ⁡ open parenthesis θ close parenthesis plus 8 sin ⁡ open parenthesis θ close parenthesis plus 7 equals 0")[$sin^(2) ( θ ) + 8 sin ( θ ) + 7 = 0$]


      
      For Problems 65–68, use Snell's Law to answer the question.


      
        A light ray passes from water to glass, with a #math.equation(block: false, alt: "19 degrees")[$19^(∘)$] angle of incidence.  What is the angle of refraction?


        
          #math.equation(block: true, alt: "17.22 degrees")[$17.22^(∘)$]


        
      
      
        A light ray passes from water to glass, with an #math.equation(block: false, alt: "82 degrees")[$82^(∘)$] angle of incidence.  What is the angle of refraction?


      
      
        A light ray passes from water to glass, with a #math.equation(block: false, alt: "32 degrees")[$32^(∘)$] angle of refraction.  What is the angle of incidence?


        
          #math.equation(block: true, alt: "35.66 degrees")[$35.66^(∘)$]


        
      
      
        A light ray passes from water to glass, with a #math.equation(block: false, alt: "58 degrees")[$58^(∘)$] angle of refraction.  What is the angle of incidence?


      
      
        + Use your calculator to graph the function #math.equation(block: false, alt: "y equals tan ⁡ open parenthesis θ close parenthesis")[$y = tan ( θ )$] in the ZTrig window (press #math.equation(block: false, alt: "ZOOM 7")[$#box(stroke: 0.6pt, inset: 3pt, outset: 0pt)[$display("ZOOM")$] " " #box(stroke: 0.6pt, inset: 3pt, outset: 0pt)[$display("7")$]$]), along with the horizontal line #math.equation(block: false, alt: "y equals 2")[$y = 2$].  Use the intersect feature to verify that the solutions of the equation #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals 2")[$tan ( θ ) = 2$]  differ by #math.equation(block: false, alt: "180 degrees")[$180^(∘)$].
+ Repeat part (a) with the horizontal line #math.equation(block: false, alt: "y equals minus 2")[$y = − 2$] to verify that the solutions of the equation #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals minus 2")[$tan ( θ ) = − 2$] differ by #math.equation(block: false, alt: "180 degrees")[$180^(∘)$].


      
      
        + What is the angle in the third quadrant with reference angle #math.equation(block: false, alt: "θ")[$θ$]? Show this angle differs from #math.equation(block: false, alt: "θ")[$θ$] by #math.equation(block: false, alt: "180 degrees")[$180^(∘)$]. Explain how this fact shows that the solutions of #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals k")[$tan ( θ ) = k$], for #math.equation(block: false, alt: "k greater than 0")[$k > 0$], differ by #math.equation(block: false, alt: "180 degrees")[$180^(∘)$].
+ What is the angle in the second quadrant with reference angle #math.equation(block: false, alt: "θ")[$θ$] ? What is the angle in the fourth quadrant with reference angle #math.equation(block: false, alt: "θ")[$θ$] ?  Show that these two angles differ by #math.equation(block: false, alt: "180 degrees")[$180^(∘)$]. Explain how this fact shows that the solutions #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals k")[$tan ( θ ) = k$], for #math.equation(block: false, alt: "k less than 0")[$k < 0$], differ by #math.equation(block: false, alt: "180 degrees")[$180^(∘)$].