5.2 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.

Note

Algebra Refresher

It is important to distinguish between an algebraic expression and an equation. An 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.

$\begin{aligned} 5 (2 + 6) & = 5 (2) + 5 (6) \\ \sqrt{3^{2} + 4^{2}} & = 3 + 4 \\ x^{2} + 3 x & = 10 \end{aligned}$

The first equation is true, the second is false, and the third equation is true only if $x equals 2$ or $x = - 5$ . When you 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.

Example 1

Evaluate $2 x + \sqrt[3]{x - 1}$ for $x equals 9$ .
Solve $2 x + \sqrt[3]{x - 1} = 20$ .

Substitute $x equals 9$ into the expression to find
$\begin{aligned} 2 x + \sqrt[3]{x - 1} & = 2 (9) + \sqrt[3]{9 - 1} \\ = 18 + \sqrt[3]{8} = 20 \end{aligned}$
We must find a value for $x$ that makes $2 x + \sqrt[3]{x - 1}$ equal to $20$ . In part (a) we saw that this expression equals $20$ when $x equals 9$ , so the solution of the equation $2 x + \sqrt[3]{x - 1} = 20$ is $x equals 9$ .

Use trial and error to find a solution of the equation

$2 x^{3} - 4 x = 5 x^{2} - 3$

Try small integer values for $x$ .

$x equals 3$

Another useful equation-solving method uses graphs.

Use a graph to show that the equation $x^{2} - 2 x + 4 = 0$ has no real-valued solutions.

The graph does not cross the line $y equals 0$ .

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.

If you are in the bottom carriage of the Ferris wheel at the start of its revolution, your height after $t$ seconds is given by

$h = f (t) = 139 - 125 \cos (\frac{2 t}{3})$

For how long are you more than 240 feet above the ground?

The figure below shows a graph of the height function and a horizontal line at $h equals 240$ .

From the graph, we see that $h equals 240$ 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 $h equals 240$ , or

$139 - 125 \cos (\frac{2 t}{3}) = 240$

To find a more precise solution, we can use algebraic methods. To start, we'll solve the slightly simpler equation

$139 - 125 \cos (θ) = 240$

We'll look for all solutions for $θ$ between $0 degrees$ and $360 degrees$ . We begin by isolating the trigonometric ratio on one side of the equation.

$\begin{aligned} 139 - 125 \cos (θ) & = 240 & Subtract 139 from both sides. \\ - 125 \cos (θ) & = 101 & Divide both sides by - 125. \\ \cos (θ) & = - 0.808 \end{aligned}$

We have solved equations like this one before: we use the inverse cosine to solve for $θ$ . Remember that there are two angles between $0 degrees$ and $360 degrees$ that have a cosine of $minus 0.808$ , one in the second quadrant and one in the third quadrant. The calculator will give us only the second quadrant solution.

$θ = \cos^{- 1} (- 0.808) = {143.9}^{\circ}$

To find the second solution, we need the third-quadrant angle whose cosine is $minus 0.808$ .

Now, the reference angle for $143.9 degrees$ is

$180^{\circ} - {143.9}^{\circ} = {36.1}^{\circ}$

and the angle in the third quadrant with the same reference angle is

$180^{\circ} + {36.1}^{\circ} = {216.1}^{\circ}$

(See the figure at right.) Thus, the other solution is $216.1 degrees$ .

To solve simple equations involving a single trigonometric ratio (either $\sin (θ), \cos (θ),$ or $\tan (θ)$ ), we can follow the steps below.

To finish up our Ferris wheel example, we can replace $θ$ by $\frac{2 t}{3}$ to get two equations:

$\begin{aligned} \frac{2 t}{3} & = 143.9 \\ 2 t & = 431.7 \\ t & = 215.35 \end{aligned}$

$\begin{aligned} \frac{2 t}{3} & = 216.1 \\ 2 t & = 648.3 \\ t & = 324.5 \end{aligned}$

Your height is above 240 feet between 215.35 seconds and 324.15 seconds. Our estimates from the graphical solutions were pretty good.

Example 3

Solve the equation $8 \sin (θ) - 1 = 3$ for $0 degrees less than or equal to θ less than or equal to 360 degrees$ .

We begin by isolating the trig ratio.

$\begin{aligned} 8 \sin (θ) - 1 & = 3 & Add 1 to both sides. \\ 8 \sin (θ) & = 4 & Divide both sides by 8. \\ \sin (θ) & = \frac{4}{8} = \frac{1}{2} \end{aligned}$

We recognize that $\sin (θ) = \frac{1}{2}$ for one of the special angles, namely $θ equals 30 degrees$ , and also for the obtuse angle $θ equals 150 degrees$ , as shown at right.

We know that $θ equals 150 degrees$ is the second-quadrant angle with reference angle $θ equals 30 degrees$ . What about the other quadrants? Because the sine is negative in the third and fourth quadrants, there are no other solutions between $0 degrees$ and $360 degrees$ . The solutions we want are $θ equals 30 degrees$ and $θ equals 150 degrees$ .

In the previous Example, we found two solutions of the equation $8 \sin (θ) - 1 = 3$ . The equation actually has infinitely many solutions, as you can see in the figure below, which shows a graph of $y = 8 \sin (θ) - 1$ and the horizontal line $y equals 3$ . The line intersects the sine graph infinitely many times, twice in each cycle.

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

Solve the equation $5 \tan (θ) + 7 = 2$ for $0 degrees less than or equal to θ less than or equal to 360 degrees$ .

We first isolate the trig ratio to get $\tan (θ) = - 1$ . We recognize the special angle $tan ⁡ open parenthesis 45 degrees close parenthesis equals 1$ . Because the tangent is negative in the second and fourth quadrants, we find the solutions $θ = 135^{\circ} and θ = 315^{\circ}$ .

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

Example 4

Solve the equation $3 \tan (β) + 1 = - 8$ for $0 degrees less than or equal to θ less than or equal to 360 degrees$ . Round your solutions to three decimal places.

First, we isolate the trig ratio.

$\begin{aligned} 3 \tan (β) + 1 & = - 8 \\ 3 \tan (β) & = - 9 \\ \tan (β) & = - 3 \end{aligned}$

There are two angles with tangent $minus 3$ , one in the second quadrant and one in the fourth quadrant. The calculator finds the solution

$\tan^{- 1} (- 3) = - {71.565}^{\circ}$

This angle is in the fourth quadrant, but it is not between $0 degrees$ and $360 degrees$ .

The angle we want is coterminal with $minus 71.565 degrees$ , so we add $360 degrees$ to get the first solution:

$B = \tan^{- 1} (- 3) + 360^{\circ} = {288.435}^{\circ}$

The other solution is in the second quadrant, as shown in the figure. This angle is $180 degrees$ less than the fourth quadrant solution:

$B = \tan^{- 1} (- 3) + 180^{\circ} = {108.435}^{\circ}$

Solve the equation $4 \cos (C) - 6 = - 3$ , for $0 degrees less than or equal to C less than or equal to 360 degrees$ . Round your solutions to three decimal places.

We isolate the trig ratio to find $\cos (C) = \frac{3}{4}$ , and $\cos^{- 1} (\frac{3}{4}) = {41.4}^{\circ}$ . The second solution lies in the fourthe quadrant at $360^{\circ} - {41.4}^{\circ} = {318.6}^{\circ}$ .

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 $minus 1$ to $1$ .

Example 5

Solve $\frac{\sin (β)}{5} - 3 = 1$ for $0 degrees less than or equal to θ less than or equal to 360 degrees$ .

We begin by isolating $\sin (β)$ .

$\begin{aligned} \frac{\sin (β)}{5} - 3 & = 1 & Add 3 to both sides. \\ \frac{\sin (β)}{5} & = 4 & Multiply both sides by 5. \\ \sin (β) & = 20 \end{aligned}$

Because $\sin (β)$ is never greater than 1, there is no angle $β$ whose sine is 20. The equation has no solution.

Solve $\cos (θ) + 5 = 3$ for $0 degrees less than or equal to θ less than or equal to 360 degrees$ .

Because $\cos (θ)$ cannot equal $minus 2$ , 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.

Use a graph to verify your solutions to $4 \cos (C) - 6 = - 3$ from Exercise.

The graphs intersect at about $40 degrees$ and about $320 degrees$ .

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

$4 x^{2} + 3 = 15$

we first isolate $x^{2}$ :

$\begin{aligned} 4 x^{2} & = 12 \\ x^{2} & = 3 \end{aligned}$

and then take square roots of both sides to find

$x = \pm \sqrt{3} \approx \pm 1.732$

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.

Example 7

Solve $4 \tan^{2} (θ) + 3 = 15$ for $0 degrees less than or equal to θ less than or equal to 360 degrees$ .

We begin by isolating the power of tangent, $\tan^{2} θ$ .

$\begin{aligned} 4 \tan^{2} (θ) & = 12 \\ \tan^{2} (θ) & = 3 \end{aligned}$

Next, we solve for $\tan (θ)$ by extracting roots.

$\tan (θ) = \pm \sqrt{3}$

There are two angles between $0 degrees$ and $360 degrees$ with tangent $the square root of 3$ and two angles with tangent $- \sqrt{3}$ , making four solutions to this equation. We know that $\tan (60^{\circ}) = \sqrt{3}$ , so one of the solutions is $60 degrees$ .

The tangent function is also positive in the third quadrant, and the angle in the third quadrant with reference angle $60 degrees$ is $180^{\circ} + 60^{\circ} = 240^{\circ}$ . The angles in the second and fourth quadrants with reference angle $60 degrees$ have tangent $- \sqrt{3}$ . These angles are

$180^{\circ} - 60^{\circ} = 120^{\circ} and 360^{\circ} - 60^{\circ} = 300^{\circ}$

The four solutions are shown at right.

Solve the equation $2 sin squared ⁡ open parenthesis θ close parenthesis equals 1$ for $0 degrees less than or equal to θ less than or equal to 360 degrees$ .

$θ equals 45 degrees$ , $θ equals 135 degrees$ , $θ equals 225 degrees$ , or $θ equals 315 degrees$

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

$4 x^{2} + 4 x - 3 = 0$

by factoring the left side to get

$(2 x + 3) (2 x - 1) = 0$

Then we apply the Zero Factor Principle to set each factor equal to zero, and solve each equation.

$\begin{aligned} 2 x + 3 & = 12 & 2 x - 1 & = 0 \\ x & = \frac{- 3}{2} & x & = \frac{1}{2} \end{aligned}$

The solutions are $\frac{- 3}{2}$ and $\frac{1}{2}$ .

Example 8

Solve $2 \cos^{2} (θ) - \cos (θ) - 1 = 0$ for $0 degrees less than or equal to θ less than or equal to 360 degrees$ .

Compare this equation to the algebraic equation $2 x^{2} - x - 1 = 0$ . Our equation has the same algebraic form, but with $x$ replaced by $\cos (θ)$ .

We can solve the algebraic equation by factoring $2 x^{2} - x - 1$ as $(2 x + 1) (x - 1),$ and we'll use the same strategy on the trigonometric equation.

$\begin{aligned} 2 \cos^{2} (θ) - \cos (θ) - 1 & = 0 & Factor the left side. \\ (2 \cos (θ) + 1) (\cos (θ) - 1) & = 0 & Set each factor equal to zero. \\ \cos (θ) = \frac{- 1}{2} or \cos (θ) & = 1 \end{aligned}$

Now we solve each equation for $θ$ . We know that $\cos (60^{\circ}) = \frac{1}{2},$ and the cosine function is negative in the second and third quadrants. The angles in those quadrants with reference angle $60 degrees$ are

$180^{\circ} - 60^{\circ} = 120^{\circ} and 180^{\circ} + 60^{\circ} = 240^{\circ}$

Also, $cos ⁡ open parenthesis θ close parenthesis equals 1$ when $θ equals 0 degrees$ . Thus, the original equation has three solutions: $θ = 120^{\circ}, θ = 240^{\circ},$ and $θ equals 0 degrees.$

Solve $9 \sin^{2} (θ) - 3 \sin (θ) = 2$ for $0 degrees less than or equal to θ less than or equal to 360 degrees$ . Round your answers to the nearest degree.

We first solve the equation $9 \sin^{2} (θ) - 3 \sin (θ) - 2 = 0$ by factoring:

$(3 \sin (θ) - 2) (3 \sin (θ) + 1) = 0$

, so $\sin (θ) = \frac{2}{3}$ or $\sin (θ) = \frac{- 1}{3}$ . From the first equation we find $θ equals 42 degrees$ and $θ equals 138 degrees$ , and from the second euqtion we find $θ equals 199 degrees$ and $θ equals 341 degrees$ .

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.

A light ray enters the boundary between the two media at a certain angle, called the angle of incidence, but leaves the boundary at a different angle, the angle of refraction. Both angles are acute angles measured from the 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:

$\frac{\sin (θ_{1})}{\sin (θ_{2})} = \frac{v_{1}}{v_{2}}$

where $θ sub 1$ is the angle in the medium where light travels at speed $v sub 1$ , and $θ sub 2$ is the angle where light travels at speed $v sub 2$ . The ratio of the speeds is called the index of refraction.

Example 9

The index of refraction from water to glass is 1.1. If light passes from water to glass with a $23 degrees$ angle of incidence, what is the angle of refraction?

If $θ$ is the angle of refraction, then from Snell's Law we have

$\begin{aligned} \frac{\sin (23^{\circ})}{\sin (θ)} & = 1.1 & Multiply both sides by \sin (θ) . \\ \sin (23^{\circ}) & = 1.1 \sin (θ) & Divide both sides by 1.1. \\ \sin (θ) & = \frac{\sin (23^{\circ})}{1.1} = 0.3552 \end{aligned}$

Because $sin ⁡ open parenthesis θ close parenthesis equals 0.3552$ , $θ = \sin^{- 1} (0.3552) = {20.8}^{\circ} .$ (For Snell's Law we use only acute angles.) The angle of refraction is approximately $20.8 degrees$ .

A light ray passes from water to glass with an $18 degrees$ angle of incidence. What is the angle of refraction?

$16.315 degrees$

Review the following skills you will need for this section.

Algebra Refresher

Each of the following "solutions" contains an error. Find the error, and supply a correct solution.

$\begin{aligned} 3 x^{2} - 5 x & = 0 \\ 3 x^{2} & = 5 x \\ 3 x & = 5 \\ x & = \frac{5}{3} \end{aligned}$
$\begin{aligned} 4 x^{2} & = 12 \\ x^{2} & = 3 \\ x & = \sqrt{3} \end{aligned}$
$\begin{aligned} (2 x + 1)^{2} & = 4 \\ 2 x + 1 & = 2 \\ 2 x & = 1 \\ x & = \frac{1}{2} \end{aligned}$
$\begin{aligned} - 2 x^{2} - 4 x & = 3 \\ - 2 x (x + 2) & = 3 \\ - 2 x = 3 & x + 2 = 3 \\ x = \frac{- 3}{2} & x = 1 \end{aligned}$

$\underset{―}{}$

Algebra Refresher Answers

We can't divide by $x . x = 0, \frac{5}{3}$
3 has two square roots. $x = \pm \sqrt{3}$
4 has two square roots. $x = \frac{- 3}{2}, \frac{1}{2}$
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 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 $0 degrees$ and $360 degrees$ .
We can use factoring or extraction of roots to solve some quadratic equations.

Study Questions

How many solutions between $0 degrees$ and $360 degrees$ does the equation $\cos (θ) = k$ have for each value of $k$ between -1 and 0?
How many solutions between $0 degrees$ and $360 degrees$ does the equation $\cos (θ) = k$ have for each value of $k$ greater than 1?
How many solutions between $0 degrees$ and $360 degrees$ does the equation $\sin^{2} (θ) = k$ have for each value of $k$ 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.)

$250 degrees$

$70 degrees$

$145 degrees$

$320 degrees$

$40 degrees$

$minus 110 degrees$

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

$18 degrees$

I: $18 degrees;$ II: $162 degrees;$ III: $198 degrees;$ IV: $342 degrees$

$35 degrees$

$52 degrees$

I: $52 degrees;$ II: $128 degrees;$ III: $232 degrees;$ IV: $308 degrees$

$78 degrees$

For Problems 9–14,

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

$x^{3} - 3 x^{2} + 4; x = - 1, 0, 1, 2, 3$
$x^{3} - 3 x^{2} + 4 = 0$

$0, 4, 2, 0, 4$
$minus 1 or 2$

$\sqrt{x} + \sqrt{2 x + 1}; x = 0, 2, 4, 6$
$\sqrt{x} + \sqrt{2 x + 1} = 5$

$\sin (θ) + \cos (θ); θ = 0^{\circ}, 30^{\circ}, 45^{\circ}, 60^{\circ}$
$\sin (θ) + \cos (θ) = \sqrt{2}$

$1, \frac{\sqrt{3} + 1}{2}, \sqrt{2}, \frac{\sqrt{3} + 1}{2}$
$45 degrees$

$\sin^{2} (α) - \cos (α); α = 45^{\circ}, 90^{\circ}, 135^{\circ}, 180^{\circ}$
$\sin^{2} (α) - \cos (α) = 1$

$\sin (β) + 2 \cos^{2} (β); β = 210^{\circ}, 225^{\circ}, 240^{\circ}, 270^{\circ}$
$\sin (β) + 2 \cos^{2} (β) = - 1$

$0, \frac{2 - \sqrt{2}}{2}, \frac{1 - \sqrt{3}}{2}, - 1$
$270 degrees$

$3 \cos^{2} (ϕ) - \sin^{2} (ϕ); ϕ = 270^{\circ}, 300^{\circ}, 315^{\circ}, 330^{\circ}$
$3 \cos^{2} (ϕ) - \sin^{2} (ϕ) = 2$

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

$\frac{- 1}{3} x^{2} + \frac{2}{3} x + 5 = 0$

$x = 5, - 3$

$0.0625 x^{2} + 0.5 x = - 1$

$x^{3} + 2 x^{2} - 6 = 2 x^{2} + 7 x$

$x = - 3, 1, 2$

$8 - 12 x + 6 x^{2} - x^{3}$

For Problems 19–32, solve the equation exactly for $0 degrees less than or equal to θ less than 360 degrees$ .

$3 \tan (θ) = \sqrt{3}$

$θ equals 30 degrees$ or $θ equals 210 degrees$

$7 \sin (θ) + 11 = 11$

$3 = 5 - 4 \cos (θ)$

$θ equals 60 degrees$ or $θ equals 300 degrees$

$6 \tan (θ) + 21 = 15$

$8 \sin (θ) + 5 = 1$

$θ equals 210 degrees$ or $θ equals 330 degrees$

$9 \cos (θ) + 15 = 6$

$0 = \sqrt{2} + 2 \sin (θ)$

$θ equals 225 degrees$ or $θ equals 315 degrees$

$\sqrt{3} \cos (θ) = - \frac{3}{2}$

$\cos^{2} (θ) - 1 = 0$

$θ equals 0 degrees$ or $θ equals 180 degrees$

$1 - \sin^{2} (θ) = 0$

$4 \sin^{2} (θ) - 3 = 0$

$θ = 60^{\circ}, θ = 120^{\circ}, θ = 240^{\circ},$ or $θ equals 300 degrees$

$0 = 1 - 2 \cos^{2} (θ)$

$1 - \tan^{2} (θ) = 0$

$θ = 45^{\circ}, θ = 135^{\circ}, θ = 225^{\circ},$ or $θ equals 315 degrees$

$0 = 6 \tan^{2} (θ) - 2$

For Problems 33–38, solve the equation for $0 degrees less than or equal to θ less than 360 degrees$ . Round your answers to two decimal places.

$\frac{1}{2} \tan (θ) - 1 = - 3$

$θ equals 104.04 degrees$ or $θ equals 284.04 degrees$

$3 \tan (θ) - 2 = 4$

$3 equals 5 cos ⁡ open parenthesis θ close parenthesis$

$θ equals 53.13 degrees$ or $θ equals 306.87 degrees$

$4 equals 6 sin ⁡ open parenthesis θ close parenthesis$

$7 \sin (θ) + 2 = 1$

$θ equals 188.21 degrees$ or $θ equals 351.79 degrees$

$2 = 5 - \frac{1}{3} \tan (θ)$

For Problems 39–46,

Use a graph to estimate the solutions for angles between $0 degrees$ and $360 degrees$ .
Solve the equation algebraically.

$7 - \tan (A) = 8$

$A equals 135 degrees$ or $A equals 315 degrees$

$6 = 8 \tan (w) - 2$

$5 = 1 - 8 \sin (ϕ)$

$ϕ equals 210 degrees$ or $ϕ equals 330 degrees$

$9 - 4 \sin (t) = 13$

$2 \cos (B) - 2 = - 2$

$B = 90^{\circ} or B = 270^{\circ}$

$2 - 6 \cos (u) = 5$

$3 = 2 \sin (θ) + 4$

$θ equals 210 degrees$ or $θ equals 330 degrees$

$5 = 3 \cos (x) + 5$

For Problems 47–52,

Use a graph to estimate the solutions for angles between $0 degrees$ and $360 degrees$ .
Solve the equation algebraically, rounding angles to the nearest degree.

$8 \sin (t) + 7 = 4$

$t equals 202 degrees$ or $t equals 338 degrees$

$9 - 6 \cos (A) = 5$

$5 \tan (B) - 4 = - 2$

$B = 22^{\circ} or B = 202^{\circ}$

$3 - 10 \tan (C) = - 11$

$1 + 6 \cos (ϕ) = - 4$

$ϕ equals 146 degrees$ or $ϕ equals 214 degrees$

$4 \sin (u) - 2 = - 1$

For Problems 53–64, solve the equation for $0 degrees less than or equal to θ less than 360 degrees$ . Round angles to two decimal places.

$6 cos squared ⁡ open parenthesis θ close parenthesis equals 2$

$θ = {54.74}^{\circ}, θ = {125.26}^{\circ}, θ = {234.74}^{\circ},$ or $θ equals 305.26 degrees$

$2 - 7 \sin^{2} (θ) = 1$

$5 \sin^{2} (θ) + \sin (θ) = 0$

$θ equals 0 degrees$ , $θ equals 180 degrees$ , $θ equals 191.54 degrees,$ or $θ equals 348.46 degrees$

$4 tan squared ⁡ open parenthesis θ close parenthesis equals tan ⁡ open parenthesis θ close parenthesis$

$2 \cos^{2} (θ) + \cos (θ) - 1 = 0$

$θ equals 60 degrees$ , $θ equals 180 degrees$ , or $θ equals 300 degrees$

$\tan^{2} (θ) - 5 \tan (θ) + 6 = 0$

$6 \tan^{2} (θ) - \tan (θ) - 1 = 0$

$θ equals 26.57 degrees$ , $θ equals 161.57 degrees$ , $θ equals 206.57 degrees$ , or $θ equals 341.57 degrees$

$10 \cos^{2} (θ) - 7 \cos (θ) + 1 = 0$

$\tan^{2} (θ) - 2 \tan (θ) = 15$

$θ equals 78.69 degrees$ , $θ equals 108.43 degrees$ , $θ equals 258.69 degrees$ , or $θ equals 288.43 degrees$

$\tan (θ) = \tan^{2} (θ) - 20$

$\cos^{2} (θ) - 4 \cos (θ) + 3 = 0$

$θ equals 0 degrees$

$\sin^{2} (θ) + 8 \sin (θ) + 7 = 0$

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

A light ray passes from water to glass, with a $19 degrees$ angle of incidence. What is the angle of refraction?

$17.22 degrees$

A light ray passes from water to glass, with an $82 degrees$ angle of incidence. What is the angle of refraction?

A light ray passes from water to glass, with a $32 degrees$ angle of refraction. What is the angle of incidence?

$35.66 degrees$

A light ray passes from water to glass, with a $58 degrees$ angle of refraction. What is the angle of incidence?

Use your calculator to graph the function $y = \tan (θ)$ in the ZTrig window (press $ZOOM 7$ ), along with the horizontal line $y equals 2$ . Use the intersect feature to verify that the solutions of the equation $tan ⁡ open parenthesis θ close parenthesis equals 2$ differ by $180 degrees$ .
Repeat part (a) with the horizontal line $y = - 2$ to verify that the solutions of the equation $\tan (θ) = - 2$ differ by $180 degrees$ .

What is the angle in the third quadrant with reference angle $θ$ ? Show this angle differs from $θ$ by $180 degrees$ . Explain how this fact shows that the solutions of $\tan (θ) = k$ , for $k greater than 0$ , differ by $180 degrees$ .
What is the angle in the second quadrant with reference angle $θ$ ? What is the angle in the fourth quadrant with reference angle $θ$ ? Show that these two angles differ by $180 degrees$ . Explain how this fact shows that the solutions $\tan (θ) = k$ , for $k less than 0$ , differ by $180 degrees$ .

Trigonometry by Katherine Yoshiwara (yoshiwarabooks.org), GNU Free Documentation License 1.2 or later. Adapted for the XYZ HTML edition with the authors' permission (recorded 2026-07-04). License: GFDL-1.2-or-later.