10.7 Trigonometric Equations and Inequalities

In Sections, and most recently, we solved some basic equations involving the trigonometric functions. Below we summarize the techniques we've employed thus far. Note that we use the neutral letter ` $u$ ' as the argument¹ of each circular function for generality.

Strategies for Solving Basic Equations Involving Trigonometric Functions

To solve $\cos (u) = c$ or $\sin (u) = c$ for $minus 1 less than or equal to c less than or equal to 1$ , first solve for $u$ in the interval $open bracket 0, 2 ⁢ π close parenthesis$ and add integer multiples of the period $2 ⁢ π$ . If $c less than minus 1$ or of $c greater than 1$ , there are no real solutions.
To solve $\sec (u) = c$ or $\csc (u) = c$ for $c less than or equal to minus 1$ or $c greater than or equal to 1$ , convert to cosine or sine, respectively, and solve as above. If $minus 1 less than c less than 1$ , there are no real solutions.
To solve $\tan (u) = c$ for any real number $c$ , first solve for $u$ in the interval $(- \frac{π}{2}, \frac{π}{2})$ and add integer multiples of the period $π$ .
To solve $\cot (u) = c$ for $c not equal to 0$ , convert to tangent and solve as above. If $c equals 0$ , the solution to $cot ⁡ open parenthesis u close parenthesis equals 0$ is $u = \frac{π}{2} + π k$ for integers $k$ .

Using the above guidelines, we can comfortably solve $\sin (x) = \frac{1}{2}$ and find the solution $x = \frac{π}{6} + 2 π k$ or $x = \frac{5 π}{6} + 2 π k$ for integers $k$ . How do we solve something like $\sin (3 x) = \frac{1}{2}$ ? Since this equation has the form $\sin (u) = \frac{1}{2}$ , we know the solutions take the form $u = \frac{π}{6} + 2 π k$ or $u = \frac{5 π}{6} + 2 π k$ for integers $k$ . Since the argument of sine here is $3 ⁢ x$ , we have $3 x = \frac{π}{6} + 2 π k$ or $3 x = \frac{5 π}{6} + 2 π k$ for integers $k$ . To solve for $x$ , we divide both sides² of these equations by $3$ , and obtain $x = \frac{π}{18} + \frac{2 π}{3} k$ or $x = \frac{5 π}{18} + \frac{2 π}{3} k$ for integers $k$ . This is the technique employed in the example below.

Example 1

Solve the following equations and check your answers analytically. List the solutions which lie in the interval $open bracket 0, 2 ⁢ π close parenthesis$ and verify them using a graphing utility.

$\cos (2 x) = - \frac{\sqrt{3}}{2}$
$\csc (\frac{1}{3} x - π) = \sqrt{2}$
$cot ⁡ open parenthesis 3 ⁢ x close parenthesis equals 0$
$sec squared ⁡ open parenthesis x close parenthesis equals 4$
$\tan (\frac{x}{2}) = - 3$
$sin ⁡ open parenthesis 2 ⁢ x close parenthesis equals 0.87$

The solutions to $\cos (u) = - \frac{\sqrt{3}}{2}$ are $u = \frac{5 π}{6} + 2 π k$ or $u = \frac{7 π}{6} + 2 π k$ for integers $k$ . Since the argument of cosine here is $2 ⁢ x$ , this means $2 x = \frac{5 π}{6} + 2 π k$ or $2 x = \frac{7 π}{6} + 2 π k$ for integers $k$ . Solving for $x$ gives $x = \frac{5 π}{12} + π k$ or $x = \frac{7 π}{12} + π k$ for integers $k$ . To check these answers analytically, we substitute them into the original equation. For any integer $k$ we have
$\begin{matrix} \cos (2 [\frac{5 π}{12} + π k]) & = & \cos (\frac{5 π}{6} + 2 π k) \\ = & \cos (\frac{5 π}{6}) & (the period of cosine is 2 π) \\ = & - \frac{\sqrt{3}}{2} \end{matrix}$
Similarly, we find $\cos (2 [\frac{7 π}{12} + π k]) = \cos (\frac{7 π}{6} + 2 π k) = \cos (\frac{7 π}{6}) = - \frac{\sqrt{3}}{2}$ . To determine which of our solutions lie in $open bracket 0, 2 ⁢ π close parenthesis$ , we substitute integer values for $k$ . The solutions we keep come from the values of $k equals 0$ and $k equals 1$ and are $x = \frac{5 π}{12}$ , $\frac{7 π}{12}$ , $\frac{17 π}{12}$ and $\frac{19 π}{12}$ . Using a calculator, we graph $y equals cos ⁡ open parenthesis 2 ⁢ x close parenthesis$ and $y = - \frac{\sqrt{3}}{2}$ over $open bracket 0, 2 ⁢ π close parenthesis$ and examine where these two graphs intersect. We see that the $x$ -coordinates of the intersection points correspond to the decimal representations of our exact answers.
Since this equation has the form $\csc (u) = \sqrt{2}$ , we rewrite this as $\sin (u) = \frac{\sqrt{2}}{2}$ and find $u = \frac{π}{4} + 2 π k$ or $u = \frac{3 π}{4} + 2 π k$ for integers $k$ . Since the argument of cosecant here is $(\frac{1}{3} x - π)$ ,
$\frac{1}{3} x - π = \frac{π}{4} + 2 π k or \frac{1}{3} x - π = \frac{3 π}{4} + 2 π k$
To solve $\frac{1}{3} x - π = \frac{π}{4} + 2 π k$ , we first add $π$ to both sides
$\frac{1}{3} x = \frac{π}{4} + 2 π k + π$
A common error is to treat the ` $2 ⁢ π ⁢ k$ ' and ` $π$ ' terms as `like' terms and try to combine them when they are not.³ We can, however, combine the ` $π$ ' and ` $\frac{π}{4}$ ' terms to get
$\frac{1}{3} x = \frac{5 π}{4} + 2 π k$
We now finish by multiplying both sides by $3$ to get
$x = 3 (\frac{5 π}{4} + 2 π k) = \frac{15 π}{4} + 6 π k$
Solving the other equation, $\frac{1}{3} x - π = \frac{3 π}{4} + 2 π k$ produces $x = \frac{21 π}{4} + 6 π k$ for integers $k$ . To check the first family of answers, we substitute, combine line terms, and simplify.
$\begin{matrix} \csc (\frac{1}{3} [\frac{15 π}{4} + 6 π k] - π) & = & \csc (\frac{5 π}{4} + 2 π k - π) \\ = & \csc (\frac{π}{4} + 2 π k) \\ = & \csc (\frac{π}{4}) & (the period of cosecant is 2 π) \\ = & \sqrt{2} \end{matrix}$
The family $x = \frac{21 π}{4} + 6 π k$ checks similarly. Despite having infinitely many solutions, we find that none of them lie in $open bracket 0, 2 ⁢ π close parenthesis$ . To verify this graphically, we use a reciprocal identity to rewrite the cosecant as a sine and we find that $y = \frac{1}{\sin (\frac{1}{3} x - π)}$ and $y = \sqrt{2}$ do not intersect at all over the interval $open bracket 0, 2 ⁢ π close parenthesis$ .
Figure 10.223
Figure 10.224
$y equals cos ⁡ open parenthesis 2 ⁢ x close parenthesis$ and $y = - \frac{\sqrt{3}}{2}$
$y = \frac{1}{\sin (\frac{1}{3} x - π)}$ and $y = \sqrt{2}$
Since $cot ⁡ open parenthesis 3 ⁢ x close parenthesis equals 0$ has the form $cot ⁡ open parenthesis u close parenthesis equals 0$ , we know $u = \frac{π}{2} + π k$ , so, in this case, $3 x = \frac{π}{2} + π k$ for integers $k$ . Solving for $x$ yields $x = \frac{π}{6} + \frac{π}{3} k$ . Checking our answers, we get
$\begin{matrix} \cot (3 [\frac{π}{6} + \frac{π}{3} k]) & = & \cot (\frac{π}{2} + π k) \\ = & \cot (\frac{π}{2}) & (the period of cotangent is π) \\ = & 0 \end{matrix}$
As $k$ runs through the integers, we obtain six answers, corresponding to $k equals 0$ through $k equals 5$ , which lie in $open bracket 0, 2 ⁢ π close parenthesis$ : $x = \frac{π}{6}$ , $\frac{π}{2}$ , $\frac{5 π}{6}$ , $\frac{7 π}{6}$ , $\frac{3 π}{2}$ and $\frac{11 π}{6}$ . To confirm these graphically, we must be careful. On many calculators, there is no function button for cotangent. We choose⁴ to use the quotient identity $\cot (3 x) = \frac{\cos (3 x)}{\sin (3 x)}$ . Graphing $y = \frac{\cos (3 x)}{\sin (3 x)}$ and $y equals 0$ (the $x$ -axis), we see that the $x$ -coordinates of the intersection points approximately match our solutions.
The complication in solving an equation like $sec squared ⁡ open parenthesis x close parenthesis equals 4$ comes not from the argument of secant, which is just $x$ , but rather, the fact the secant is being squared. To get this equation to look like one of the forms listed on page, we extract square roots to get $\sec (x) = \pm 2$ . Converting to cosines, we have $\cos (x) = \pm \frac{1}{2}$ . For $\cos (x) = \frac{1}{2}$ , we get $x = \frac{π}{3} + 2 π k$ or $x = \frac{5 π}{3} + 2 π k$ for integers $k$ . For $\cos (x) = - \frac{1}{2}$ , we get $x = \frac{2 π}{3} + 2 π k$ or $x = \frac{4 π}{3} + 2 π k$ for integers $k$ . If we take a step back and think of these families of solutions geometrically, we see we are finding the measures of all angles with a reference angle of $\frac{π}{3}$ . As a result, these solutions can be combined and we may write our solutions as $x = \frac{π}{3} + π k$ and $x = \frac{2 π}{3} + π k$ for integers $k$ . To check the first family of solutions, we note that, depending on the integer $k$ , $\sec (\frac{π}{3} + π k)$ doesn't always equal $\sec (\frac{π}{3})$ . However, it is true that for all integers $k$ , $\sec (\frac{π}{3} + π k) = \pm \sec (\frac{π}{3}) = \pm 2$ . (Can you show this?) As a result,
$\begin{matrix} \sec^{2} (\frac{π}{3} + π k) & = & {(\pm \sec (\frac{π}{3}))}^{2} \\ = & {(\pm 2)}^{2} \\ = & 4 \end{matrix}$
The same holds for the family $x = \frac{2 π}{3} + π k$ . The solutions which lie in $open bracket 0, 2 ⁢ π close parenthesis$ come from the values $k equals 0$ and $k equals 1$ , namely $x = \frac{π}{3}$ , $\frac{2 π}{3}$ , $\frac{4 π}{3}$ and $\frac{5 π}{3}$ . To confirm graphically, we use a reciprocal identity to rewrite the secant as cosine. The $x$ -coordinates of the intersection points of $y = \frac{1}{{(\cos (x))}^{2}}$ and $y equals 4$ verify our answers.
Figure 10.225
Figure 10.226
$y = \frac{\cos (3 x)}{\sin (3 x)}$ and $y equals 0$
$y = \frac{1}{\cos^{2} (x)}$ and $y equals 4$
The equation $\tan (\frac{x}{2}) = - 3$ has the form $\tan (u) = - 3$ , whose solution is $u = \arctan (- 3) + π k$ . Hence, $\frac{x}{2} = \arctan (- 3) + π k$ , so $x = 2 \arctan (- 3) + 2 π k$ for integers $k$ . To check, we note
$tan open parenthesis the fraction 2 arctan open parenthesis minus 3 close parenthesis plus 2 π k over 2 close parenthesis, equals, tan open parenthesis arctan open parenthesis minus 3 close parenthesis plus π k close parenthesis; open bracket 3pt close bracket, equals, tan open parenthesis arctan open parenthesis minus 3 close parenthesis close parenthesis, (the period of tangent is $\pi$); open bracket 3pt close bracket, equals, minus 3, open parenthesis See Theorem \ref a r c t a n g e n t c o t a n g e n t f u n c t i o n p r o p s close parenthesis$
To determine which of our answers lie in the interval $open bracket 0, 2 ⁢ π close parenthesis$ , we first need to get an idea of the value of $2 ⁢ arctan ⁡ open parenthesis minus 3 close parenthesis$ . While we could easily find an approximation using a calculator,⁵ we proceed analytically. Since $minus 3 less than 0$ , it follows that $- \frac{π}{2} < \arctan (- 3) < 0$ . Multiplying through by $2$ gives $- π < 2 \arctan (- 3) < 0$ . We are now in a position to argue which of the solutions $x = 2 \arctan (- 3) + 2 π k$ lie in $open bracket 0, 2 ⁢ π close parenthesis$ . For $k equals 0$ , we get $x = 2 \arctan (- 3) < 0$ , so we discard this answer and all answers $x = 2 \arctan (- 3) + 2 π k$ where $k less than 0$ . Next, we turn our attention to $k equals 1$ and get $x = 2 \arctan (- 3) + 2 π$ . Starting with the inequality $- π < 2 \arctan (- 3) < 0$ , we add $2 ⁢ π$ and get $π < 2 \arctan (- 3) + 2 π < 2 π$ . This means $x = 2 \arctan (- 3) + 2 π$ lies in $open bracket 0, 2 ⁢ π close parenthesis$ . Advancing $k$ to $2$ produces $x = 2 \arctan (- 3) + 4 π$ . Once again, we get from $- π < 2 \arctan (- 3) < 0$ that $3 π < 2 \arctan (- 3) + 4 π < 4 π$ . Since this is outside the interval $open bracket 0, 2 ⁢ π close parenthesis$ , we discard $x = 2 \arctan (- 3) + 4 π$ and all solutions of the form $x = 2 \arctan (- 3) + 2 π k$ for $k greater than 2$ . Graphically, we see $y = \tan (\frac{x}{2})$ and $y = - 3$ intersect only once on $open bracket 0, 2 ⁢ π close parenthesis$ at $x = 2 \arctan (- 3) + 2 π \approx 3.7851$ .
To solve $sin ⁡ open parenthesis 2 ⁢ x close parenthesis equals 0.87$ , we first note that it has the form $sin ⁡ open parenthesis u close parenthesis equals 0.87$ , which has the family of solutions $u = \arcsin (0.87) + 2 π k$ or $u = π - \arcsin (0.87) + 2 π k$ for integers $k$ . Since the argument of sine here is $2 ⁢ x$ , we get $2 x = \arcsin (0.87) + 2 π k$ or $2 x = π - \arcsin (0.87) + 2 π k$ which gives $x = \frac{1}{2} \arcsin (0.87) + π k$ or $x = \frac{π}{2} - \frac{1}{2} \arcsin (0.87) + π k$ for integers $k$ . To check,
$sin open parenthesis 2 open bracket the fraction 1 over 2 arcsin open parenthesis 0.87 close parenthesis plus π k close bracket close parenthesis, equals, sin open parenthesis arcsin open parenthesis 0.87 close parenthesis plus 2 π k close parenthesis; open bracket 3pt close bracket, equals, sin open parenthesis arcsin open parenthesis 0.87 close parenthesis close parenthesis, (the period of sine is $2\pi$); open bracket 3pt close bracket, equals, 0.87, open parenthesis See Theorem \ref a r c c o s i n e s i n e f u n c t i o n p r o p s close parenthesis$
For the family $x = \frac{π}{2} - \frac{1}{2} \arcsin (0.87) + π k$ , we get
$sin open parenthesis 2 open bracket the fraction π over 2 minus the fraction 1 over 2 arcsin open parenthesis 0.87 close parenthesis plus π k close bracket close parenthesis, equals, sin open parenthesis π minus arcsin open parenthesis 0.87 close parenthesis plus 2 π k close parenthesis; open bracket 3pt close bracket, equals, sin open parenthesis π minus arcsin open parenthesis 0.87 close parenthesis close parenthesis, (the period of sine is $2\pi$); open bracket 3pt close bracket, equals, sin open parenthesis arcsin open parenthesis 0.87 close parenthesis close parenthesis, ($\sin(\pi - t) = \sin(t)$); open bracket 3pt close bracket, equals, 0.87, open parenthesis See Theorem \ref a r c c o s i n e s i n e f u n c t i o n p r o p s close parenthesis$
To determine which of these solutions lie in $open bracket 0, 2 ⁢ π close parenthesis$ , we first need to get an idea of the value of $x = \frac{1}{2} \arcsin (0.87)$ . Once again, we could use the calculator, but we adopt an analytic route here. By definition, $0 < \arcsin (0.87) < \frac{π}{2}$ so that multiplying through by $\frac{1}{2}$ gives us $0 < \frac{1}{2} \arcsin (0.87) < \frac{π}{4}$ . Starting with the family of solutions $x = \frac{1}{2} \arcsin (0.87) + π k$ , we use the same kind of arguments as in our solution to number above and find only the solutions corresponding to $k equals 0$ and $k equals 1$ lie in $open bracket 0, 2 ⁢ π close parenthesis$ : $x = \frac{1}{2} \arcsin (0.87)$ and $x = \frac{1}{2} \arcsin (0.87) + π$ . Next, we move to the family $x = \frac{π}{2} - \frac{1}{2} \arcsin (0.87) + π k$ for integers $k$ . Here, we need to get a better estimate of $\frac{π}{2} - \frac{1}{2} \arcsin (0.87)$ . From the inequality $0 < \frac{1}{2} \arcsin (0.87) < \frac{π}{4}$ , we first multiply through by $minus 1$ and then add $\frac{π}{2}$ to get $\frac{π}{2} > \frac{π}{2} - \frac{1}{2} \arcsin (0.87) > \frac{π}{4}$ , or $\frac{π}{4} < \frac{π}{2} - \frac{1}{2} \arcsin (0.87) < \frac{π}{2}$ . Proceeding with the usual arguments, we find the only solutions which lie in $open bracket 0, 2 ⁢ π close parenthesis$ correspond to $k equals 0$ and $k equals 1$ , namely $x = \frac{π}{2} - \frac{1}{2} \arcsin (0.87)$ and $x = \frac{3 π}{2} - \frac{1}{2} \arcsin (0.87)$ . All told, we have found four solutions to $sin ⁡ open parenthesis 2 ⁢ x close parenthesis equals 0.87$ in $open bracket 0, 2 ⁢ π close parenthesis$ : $x = \frac{1}{2} \arcsin (0.87)$ , $x = \frac{1}{2} \arcsin (0.87) + π$ , $x = \frac{π}{2} - \frac{1}{2} \arcsin (0.87)$ and $x = \frac{3 π}{2} - \frac{1}{2} \arcsin (0.87)$ . By graphing $y equals sin ⁡ open parenthesis 2 ⁢ x close parenthesis$ and $y equals 0.87$ , we confirm our results.
.5in
Figure 10.227
Figure 10.228
$y = \tan (\frac{x}{2})$ and $y = - 3$
$y equals sin ⁡ open parenthesis 2 ⁢ x close parenthesis$ and $y equals 0.87$

Each of the problems in Example Example 1 featured one trigonometric function. If an equation involves two different trigonometric functions or if the equation contains the same trigonometric function but with different arguments, we will need to use identities and Algebra to reduce the equation to the same form as those given on page.

Example 2

Solve the following equations and list the solutions which lie in the interval $open bracket 0, 2 ⁢ π close parenthesis$ . Verify your solutions on $open bracket 0, 2 ⁢ π close parenthesis$ graphically.

$3 ⁢ sin cubed ⁡ open parenthesis x close parenthesis equals sin squared ⁡ open parenthesis x close parenthesis$
$\sec^{2} (x) = \tan (x) + 3$
$\cos (2 x) = 3 \cos (x) - 2$
$\cos (3 x) = 2 - \cos (x)$
$cos ⁡ open parenthesis 3 ⁢ x close parenthesis equals cos ⁡ open parenthesis 5 ⁢ x close parenthesis$
$\sin (2 x) = \sqrt{3} \cos (x)$
$\sin (x) \cos (\frac{x}{2}) + \cos (x) \sin (\frac{x}{2}) = 1$
$\cos (x) - \sqrt{3} \sin (x) = 2$

We resist the temptation to divide both sides of $3 ⁢ sin cubed ⁡ open parenthesis x close parenthesis equals sin squared ⁡ open parenthesis x close parenthesis$ by $\sin^{2} (x)$ (What goes wrong if you do?) and instead gather all of the terms to one side of the equation and factor.
$\begin{matrix} 3 \sin^{3} (x) & = & \sin^{2} (x) \\ 3 \sin^{3} (x) - \sin^{2} (x) & = & 0 \\ \sin^{2} (x) (3 \sin (x) - 1) & = & 0 & Factor out \sin^{2} (x) from both terms. \end{matrix}$
We get $sin squared ⁡ open parenthesis x close parenthesis equals 0$ or $3 \sin (x) - 1 = 0$ . Solving for $\sin (x)$ , we find $sin ⁡ open parenthesis x close parenthesis equals 0$ or $\sin (x) = \frac{1}{3}$ . The solution to the first equation is $x = π k$ , with $x equals 0$ and $x = π$ being the two solutions which lie in $open bracket 0, 2 ⁢ π close parenthesis$ . To solve $\sin (x) = \frac{1}{3}$ , we use the arcsine function to get $x = \arcsin (\frac{1}{3}) + 2 π k$ or $x = π - \arcsin (\frac{1}{3}) + 2 π k$ for integers $k$ . We find the two solutions here which lie in $open bracket 0, 2 ⁢ π close parenthesis$ to be $x = \arcsin (\frac{1}{3})$ and $x = π - \arcsin (\frac{1}{3})$ . To check graphically, we plot $y equals 3 ⁢ open parenthesis sin ⁡ open parenthesis x close parenthesis close parenthesis cubed$ and $y = {(\sin (x))}^{2}$ and find the $x$ -coordinates of the intersection points of these two curves. Some extra zooming is required near $x equals 0$ and $x = π$ to verify that these two curves do in fact intersect four times.⁶
Analysis of $\sec^{2} (x) = \tan (x) + 3$ reveals two different trigonometric functions, so an identity is in order. Since $\sec^{2} (x) = 1 + \tan^{2} (x)$ , we get
$\begin{matrix} \sec^{2} (x) & = & \tan (x) + 3 \\ 1 + \tan^{2} (x) & = & \tan (x) + 3 & (Since \sec^{2} (x) = 1 + \tan^{2} (x) .) \\ \tan^{2} (x) - \tan (x) - 2 & = & 0 \\ u^{2} - u - 2 & = & 0 & Let u = \tan (x) . \\ (u + 1) (u - 2) & = & 0 \end{matrix}$
This gives $u = - 1$ or $u equals 2$ . Since $u = \tan (x)$ , we have $\tan (x) = - 1$ or $tan ⁡ open parenthesis x close parenthesis equals 2$ . From $\tan (x) = - 1$ , we get $x = - \frac{π}{4} + π k$ for integers $k$ . To solve $tan ⁡ open parenthesis x close parenthesis equals 2$ , we employ the arctangent function and get $x = \arctan (2) + π k$ for integers $k$ . From the first set of solutions, we get $x = \frac{3 π}{4}$ and $x = \frac{7 π}{4}$ as our answers which lie in $open bracket 0, 2 ⁢ π close parenthesis$ . Using the same sort of argument we saw in Example Example 1, we get $x equals arctan ⁡ open parenthesis 2 close parenthesis$ and $x = π + \arctan (2)$ as answers from our second set of solutions which lie in $open bracket 0, 2 ⁢ π close parenthesis$ . Using a reciprocal identity, we rewrite the secant as a cosine and graph $y = \frac{1}{{(\cos (x))}^{2}}$ and $y = \tan (x) + 3$ to find the $x$ -values of the points where they intersect.
Figure 10.229
Figure 10.230
$y equals 3 ⁢ open parenthesis sin ⁡ open parenthesis x close parenthesis close parenthesis cubed$ and $y = {(\sin (x))}^{2}$
$y = \frac{1}{{(\cos (x))}^{2}}$ and $y = \tan (x) + 3$
In the equation $\cos (2 x) = 3 \cos (x) - 2$ , we have the same circular function, namely cosine, on both sides but the arguments differ. Using the identity $\cos (2 x) = 2 \cos^{2} (x) - 1$ , we obtain a `quadratic in disguise' and proceed as we have done in the past.
$\begin{matrix} \cos (2 x) & = & 3 \cos (x) - 2 \\ 2 \cos^{2} (x) - 1 & = & 3 \cos (x) - 2 & (Since \cos (2 x) = 2 \cos^{2} (x) - 1 .) \\ 2 \cos^{2} (x) - 3 \cos (x) + 1 & = & 0 \\ 2 u^{2} - 3 u + 1 & = & 0 & Let u = \cos (x) . \\ (2 u - 1) (u - 1) & = & 0 \end{matrix}$
This gives $u = \frac{1}{2}$ or $u equals 1$ . Since $u = \cos (x)$ , we get $\cos (x) = \frac{1}{2}$ or $cos ⁡ open parenthesis x close parenthesis equals 1$ . Solving $\cos (x) = \frac{1}{2}$ , we get $x = \frac{π}{3} + 2 π k$ or $x = \frac{5 π}{3} + 2 π k$ for integers $k$ . From $cos ⁡ open parenthesis x close parenthesis equals 1$ , we get $x equals 2 ⁢ π ⁢ k$ for integers $k$ . The answers which lie in $open bracket 0, 2 ⁢ π close parenthesis$ are $x equals 0$ , $\frac{π}{3}$ , and $\frac{5 π}{3}$ . Graphing $y equals cos ⁡ open parenthesis 2 ⁢ x close parenthesis$ and $y = 3 \cos (x) - 2$ , we find, after a little extra effort, that the curves intersect in three places on $open bracket 0, 2 ⁢ π close parenthesis$ , and the $x$ -coordinates of these points confirm our results.
To solve $\cos (3 x) = 2 - \cos (x)$ , we use the same technique as in the previous problem. From Example, number, we know that $\cos (3 x) = 4 \cos^{3} (x) - 3 \cos (x)$ . This transforms the equation into a polynomial in terms of $\cos (x)$ .
$\begin{matrix} \cos (3 x) & = & 2 - \cos (x) \\ 4 \cos^{3} (x) - 3 \cos (x) & = & 2 - \cos (x) \\ 2 \cos^{3} (x) - 2 \cos (x) - 2 & = & 0 \\ 4 u^{3} - 2 u - 2 & = & 0 & Let u = \cos (x) . \end{matrix}$
To solve $4 u^{3} - 2 u - 2 = 0$ , we need the techniques in Chapter to factor $4 u^{3} - 2 u - 2$ into $(u - 1) (4 u^{2} + 4 u + 2)$ . We get either $u - 1 = 0$ or $4 u^{2} + 2 u + 2 = 0$ , and since the discriminant of the latter is negative, the only real solution to $4 u^{3} - 2 u - 2 = 0$ is $u equals 1$ . Since $u = \cos (x)$ , we get $cos ⁡ open parenthesis x close parenthesis equals 1$ , so $x equals 2 ⁢ π ⁢ k$ for integers $k$ . The only solution which lies in $open bracket 0, 2 ⁢ π close parenthesis$ is $x equals 0$ . Graphing $y equals cos ⁡ open parenthesis 3 ⁢ x close parenthesis$ and $y = 2 - \cos (x)$ on the same set of axes over $open bracket 0, 2 ⁢ π close parenthesis$ shows that the graphs intersect at what appears to be $open parenthesis 0, 1 close parenthesis$ , as required.
Figure 10.231
Figure 10.232
$y equals cos ⁡ open parenthesis 2 ⁢ x close parenthesis$ and $y = 3 \cos (x) - 2$
$y equals cos ⁡ open parenthesis 3 ⁢ x close parenthesis$ and $y = 2 - \cos (x)$
While we could approach $cos ⁡ open parenthesis 3 ⁢ x close parenthesis equals cos ⁡ open parenthesis 5 ⁢ x close parenthesis$ in the same manner as we did the previous two problems, we choose instead to showcase the utility of the Sum to Product Identities. From $cos ⁡ open parenthesis 3 ⁢ x close parenthesis equals cos ⁡ open parenthesis 5 ⁢ x close parenthesis$ , we get $\cos (5 x) - \cos (3 x) = 0$ , and it is the presence of $0$ on the right hand side that indicates a switch to a product would be a good move.⁷ Using Theorem, we have that $\cos (5 x) - \cos (3 x) = - 2 \sin (\frac{5 x + 3 x}{2}) \sin (\frac{5 x - 3 x}{2}) = - 2 \sin (4 x) \sin (x)$ . Hence, the equation $cos ⁡ open parenthesis 5 ⁢ x close parenthesis equals cos ⁡ open parenthesis 3 ⁢ x close parenthesis$ is equivalent to $- 2 \sin (4 x) \sin (x) = 0$ . From this, we get $sin ⁡ open parenthesis 4 ⁢ x close parenthesis equals 0$ or $\sin (x)$ = 0. Solving $sin ⁡ open parenthesis 4 ⁢ x close parenthesis equals 0$ gives $x = \frac{π}{4} k$ for integers $k$ , and the solution to $sin ⁡ open parenthesis x close parenthesis equals 0$ is $x = π k$ for integers $k$ . The second set of solutions is contained in the first set of solutions,⁸ so our final solution to $cos ⁡ open parenthesis 5 ⁢ x close parenthesis equals cos ⁡ open parenthesis 3 ⁢ x close parenthesis$ is $x = \frac{π}{4} k$ for integers $k$ . There are eight of these answers which lie in $open bracket 0, 2 ⁢ π close parenthesis$ : $x equals 0$ , $\frac{π}{4}$ , $\frac{π}{2}$ , $\frac{3 π}{4}$ , $π$ , $\frac{5 π}{4}$ , $\frac{3 π}{2}$ and $\frac{7 π}{4}$ . Our plot of the graphs of $y equals cos ⁡ open parenthesis 3 ⁢ x close parenthesis$ and $y equals cos ⁡ open parenthesis 5 ⁢ x close parenthesis$ below (after some careful zooming) bears this out.
In examining the equation $\sin (2 x) = \sqrt{3} \cos (x)$ , not only do we have different circular functions involved, namely sine and cosine, we also have different arguments to contend with, namely $2 ⁢ x$ and $x$ . Using the identity $sin ⁡ open parenthesis 2 ⁢ x close parenthesis equals 2 ⁢ sin ⁡ open parenthesis x close parenthesis ⁢ cos ⁡ open parenthesis x close parenthesis$ makes all of the arguments the same and we proceed as we would solving any nonlinear equation – gather all of the nonzero terms on one side of the equation and factor.
$\begin{matrix} \sin (2 x) & = & \sqrt{3} \cos (x) \\ 2 \sin (x) \cos (x) & = & \sqrt{3} \cos (x) & (Since \sin (2 x) = 2 \sin (x) \cos (x) .) \\ 2 \sin (x) \cos (x) - \sqrt{3} \cos (x) & = & 0 \\ \cos (x) (2 \sin (x) - \sqrt{3}) & = & 0 \end{matrix}$
from which we get $cos ⁡ open parenthesis x close parenthesis equals 0$ or $\sin (x) = \frac{\sqrt{3}}{2}$ . From $cos ⁡ open parenthesis x close parenthesis equals 0$ , we obtain $x = \frac{π}{2} + π k$ for integers $k$ . From $\sin (x) = \frac{\sqrt{3}}{2}$ , we get $x = \frac{π}{3} + 2 π k$ or $x = \frac{2 π}{3} + 2 π k$ for integers $k$ . The answers which lie in $open bracket 0, 2 ⁢ π close parenthesis$ are $x = \frac{π}{2}$ , $\frac{3 π}{2}$ , $\frac{π}{3}$ and $\frac{2 π}{3}$ . We graph $y equals sin ⁡ open parenthesis 2 ⁢ x close parenthesis$ and $y = \sqrt{3} \cos (x)$ and, after some careful zooming, verify our answers.
Figure 10.233
Figure 10.234
$y equals cos ⁡ open parenthesis 3 ⁢ x close parenthesis$ and $y equals cos ⁡ open parenthesis 5 ⁢ x close parenthesis$
$y equals sin ⁡ open parenthesis 2 ⁢ x close parenthesis$ and $y = \sqrt{3} \cos (x)$
Unlike the previous problem, there seems to be no quick way to get the circular functions or their arguments to match in the equation $\sin (x) \cos (\frac{x}{2}) + \cos (x) \sin (\frac{x}{2}) = 1$ . If we stare at it long enough, however, we realize that the left hand side is the expanded form of the sum formula for $\sin (x + \frac{x}{2})$ . Hence, our original equation is equivalent to $\sin (\frac{3}{2} x) = 1$ . Solving, we find $x = \frac{π}{3} + \frac{4 π}{3} k$ for integers $k$ . Two of these solutions lie in $open bracket 0, 2 ⁢ π close parenthesis$ : $x = \frac{π}{3}$ and $x = \frac{5 π}{3}$ . Graphing $y = \sin (x) \cos (\frac{x}{2}) + \cos (x) \sin (\frac{x}{2})$ and $y equals 1$ validates our solutions.
With the absence of double angles or squares, there doesn't seem to be much we can do. However, since the arguments of the cosine and sine are the same, we can rewrite the left hand side of this equation as a sinusoid.⁹ To fit $f (x) = \cos (x) - \sqrt{3} \sin (x)$ to the form $A \sin (ω t + ϕ) + B$ , we use what we learned in Example and find $A equals 2$ , $B equals 0$ , $ω equals 1$ and $ϕ = \frac{5 π}{6}$ . Hence, we can rewrite the equation $\cos (x) - \sqrt{3} \sin (x) = 2$ as $2 \sin (x + \frac{5 π}{6}) = 2$ , or $\sin (x + \frac{5 π}{6}) = 1$ . Solving the latter, we get $x = - \frac{π}{3} + 2 π k$ for integers $k$ . Only one of these solutions, $x = \frac{5 π}{3}$ , which corresponds to $k equals 1$ , lies in $open bracket 0, 2 ⁢ π close parenthesis$ . Geometrically, we see that $y = \cos (x) - \sqrt{3} \sin (x)$ and $y equals 2$ intersect just once, supporting our answer.
Figure 10.235
Figure 10.236
$y = \sin (x) \cos (\frac{x}{2}) + \cos (x) \sin (\frac{x}{2})$ and $y equals 1$
$y = \cos (x) - \sqrt{3} \sin (x)$ and $y equals 2$

We repeat here the advice given when solving systems of nonlinear equations in section – when it comes to solving equations involving the trigonometric functions, it helps to just try something.

Next, we focus on solving inequalities involving the trigonometric functions. Since these functions are continuous on their domains, we may use the sign diagram technique we've used in the past to solve the inequalities.¹⁰

Example 3

Solve the following inequalities on $open bracket 0, 2 ⁢ π close parenthesis$ . Express your answers using interval notation and verify your answers graphically.

$2 ⁢ sin ⁡ open parenthesis x close parenthesis less than or equal to 1$
$sin ⁡ open parenthesis 2 ⁢ x close parenthesis greater than cos ⁡ open parenthesis x close parenthesis$
$tan ⁡ open parenthesis x close parenthesis greater than or equal to 3$

We begin solving $2 ⁢ sin ⁡ open parenthesis x close parenthesis less than or equal to 1$ by collecting all of the terms on one side of the equation and zero on the other to get $2 ⁢ sin ⁡ open parenthesis x close parenthesis minus 1 less than or equal to 0$ . Next, we let $f (x) = 2 \sin (x) - 1$ and note that our original inequality is equivalent to solving $f ⁢ open parenthesis x close parenthesis less than or equal to 0$ . We now look to see where, if ever, $f$ is undefined and where $f ⁢ open parenthesis x close parenthesis equals 0$ . Since the domain of $f$ is all real numbers, we can immediately set about finding the zeros of $f$ . Solving $f ⁢ open parenthesis x close parenthesis equals 0$ , we have $2 \sin (x) - 1 = 0$ or $\sin (x) = \frac{1}{2}$ . The solutions here are $x = \frac{π}{6} + 2 π k$ and $x = \frac{5 π}{6} + 2 π k$ for integers $k$ . Since we are restricting our attention to $open bracket 0, 2 ⁢ π close parenthesis$ , only $x = \frac{π}{6}$ and $x = \frac{5 π}{6}$ are of concern to us. Next, we choose test values in $open bracket 0, 2 ⁢ π close parenthesis$ other than the zeros and determine if $f$ is positive or negative there. For $x equals 0$ we have $f (0) = - 1$ , for $x = \frac{π}{2}$ we get $f (\frac{π}{2}) = 1$ and for $x = π$ we get $f (π) = - 1$ . Since our original inequality is equivalent to $f ⁢ open parenthesis x close parenthesis less than or equal to 0$ , we are looking for where the function is negative $(-)$ or $0$ , and we get the intervals $[0, \frac{π}{6}] \cup [\frac{5 π}{6}, 2 π)$ . We can confirm our answer graphically by seeing where the graph of $y equals 2 ⁢ sin ⁡ open parenthesis x close parenthesis$ crosses or is below the graph of $y equals 1$ .
Figure 10.237
Figure 10.238
$y equals 2 ⁢ sin ⁡ open parenthesis x close parenthesis$ and $y equals 1$
We first rewrite $sin ⁡ open parenthesis 2 ⁢ x close parenthesis greater than cos ⁡ open parenthesis x close parenthesis$ as $sin ⁡ open parenthesis 2 ⁢ x close parenthesis minus cos ⁡ open parenthesis x close parenthesis greater than 0$ and let $f (x) = \sin (2 x) - \cos (x)$ . Our original inequality is thus equivalent to $f ⁢ open parenthesis x close parenthesis greater than 0$ . The domain of $f$ is all real numbers, so we can advance to finding the zeros of $f$ . Setting $f ⁢ open parenthesis x close parenthesis equals 0$ yields $\sin (2 x) - \cos (x) = 0$ , which, by way of the double angle identity for sine, becomes $2 \sin (x) \cos (x) - \cos (x) = 0$ or $\cos (x) (2 \sin (x) - 1) = 0$ . From $cos ⁡ open parenthesis x close parenthesis equals 0$ , we get $x = \frac{π}{2} + π k$ for integers $k$ of which only $x = \frac{π}{2}$ and $x = \frac{3 π}{2}$ lie in $open bracket 0, 2 ⁢ π close parenthesis$ . For $2 \sin (x) - 1 = 0$ , we get $\sin (x) = \frac{1}{2}$ which gives $x = \frac{π}{6} + 2 π k$ or $x = \frac{5 π}{6} + 2 π k$ for integers $k$ . Of those, only $x = \frac{π}{6}$ and $x = \frac{5 π}{6}$ lie in $open bracket 0, 2 ⁢ π close parenthesis$ . Next, we choose our test values. For $x equals 0$ we find $f (0) = - 1$ ; when $x = \frac{π}{4}$ we get $f (\frac{π}{4}) = 1 - \frac{\sqrt{2}}{2} = \frac{2 - \sqrt{2}}{2}$ ; for $x = \frac{3 π}{4}$ we get $f (\frac{3 π}{4}) = - 1 + \frac{\sqrt{2}}{2} = \frac{\sqrt{2} - 2}{2}$ ; when $x = π$ we have $f ⁢ open parenthesis π close parenthesis equals 1$ , and lastly, for $x = \frac{7 π}{4}$ we get $f (\frac{7 π}{4}) = - 1 - \frac{\sqrt{2}}{2} = \frac{- 2 - \sqrt{2}}{2}$ . We see $f ⁢ open parenthesis x close parenthesis greater than 0$ on $(\frac{π}{6}, \frac{π}{2}) \cup (\frac{5 π}{6}, \frac{3 π}{2})$ , so this is our answer. We can use the calculator to check that the graph of $y equals sin ⁡ open parenthesis 2 ⁢ x close parenthesis$ is indeed above the graph of $y = \cos (x)$ on those intervals.
Figure 10.239
Figure 10.240
$y equals sin ⁡ open parenthesis 2 ⁢ x close parenthesis$ and $y = \cos (x)$
Proceeding as in the last two problems, we rewrite $tan ⁡ open parenthesis x close parenthesis greater than or equal to 3$ as $tan ⁡ open parenthesis x close parenthesis minus 3 greater than or equal to 0$ and let $f (x) = \tan (x) - 3$ . We note that on $open bracket 0, 2 ⁢ π close parenthesis$ , $f$ is undefined at $x = \frac{π}{2}$ and $\frac{3 π}{2}$ , so those values will need the usual disclaimer on the sign diagram.¹¹ Moving along to zeros, solving $f (x) = \tan (x) - 3 = 0$ requires the arctangent function. We find $x = \arctan (3) + π k$ for integers $k$ and of these, only $x equals arctan ⁡ open parenthesis 3 close parenthesis$ and $x = \arctan (3) + π$ lie in $open bracket 0, 2 ⁢ π close parenthesis$ . Since $3 greater than 0$ , we know $0 < \arctan (3) < \frac{π}{2}$ which allows us to position these zeros correctly on the sign diagram. To choose test values, we begin with $x equals 0$ and find $f (0) = - 3$ . Finding a convenient test value in the interval $(\arctan (3), \frac{π}{2})$ is a bit more challenging. Keep in mind that the arctangent function is increasing and is bounded above by $\frac{π}{2}$ . This means that the number $x equals arctan ⁡ open parenthesis 117 close parenthesis$ is guaranteed¹² to lie between $arctan ⁡ open parenthesis 3 close parenthesis$ and $\frac{π}{2}$ . We see that $f (\arctan (117)) = \tan (\arctan (117)) - 3 = 114$ . For our next test value, we take $x = π$ and find $f (π) = - 3$ . To find our next test value, we note that since $\arctan (3) < \arctan (117) < \frac{π}{2}$ , it follows¹³ that $\arctan (3) + π < \arctan (117) + π < \frac{3 π}{2}$ . Evaluating $f$ at $x = \arctan (117) + π$ yields $f (\arctan (117) + π) = \tan (\arctan (117) + π) - 3 = \tan (\arctan (117)) - 3 = 114$ . We choose our last test value to be $x = \frac{7 π}{4}$ and find $f (\frac{7 π}{4}) = - 4$ . Since we want $f ⁢ open parenthesis x close parenthesis greater than or equal to 0$ , we see that our answer is $[\arctan (3), \frac{π}{2}) \cup [\arctan (3) + π, \frac{3 π}{2})$ . Using the graphs of $y = \tan (x)$ and $y equals 3$ , we see when the graph of the former is above (or meets) the graph of the latter.
Figure 10.241
Figure 10.242
$y = \tan (x)$ and $y equals 3$

Our next example puts solving equations and inequalities to good use – finding domains of functions.

Example 4

Express the domain of the following functions using extended interval notation.¹⁴

$f (x) = \csc (2 x + \frac{π}{3})$
$f (x) = \frac{\sin (x)}{2 \cos (x) - 1}$
$f (x) = \sqrt{1 - \cot (x)}$

To find the domain of $f (x) = \csc (2 x + \frac{π}{3})$ , we rewrite $f$ in terms of sine as $f (x) = \frac{1}{\sin (2 x + \frac{π}{3})}$ . Since the sine function is defined everywhere, our only concern comes from zeros in the denominator. Solving $\sin (2 x + \frac{π}{3}) = 0$ , we get $x = - \frac{π}{6} + \frac{π}{2} k$ for integers $k$ . In set-builder notation, our domain is ${x : x \neq - \frac{π}{6} + \frac{π}{2} k for integers k}$ . To help visualize the domain, we follow the old mantra `When in doubt, write it out!' We get ${x : x \neq - \frac{π}{6}, \frac{2 π}{6}, - \frac{4 π}{6}, \frac{5 π}{6}, - \frac{7 π}{6}, \frac{8 π}{6}, \dots}$ , where we have kept the denominators $6$ throughout to help see the pattern. Graphing the situation on a numberline, we have
Figure 10.243
Proceeding as we did on page in Section, we let $x_{k}$ denote the $k$ th number excluded from the domain and we have $x_{k} = - \frac{π}{6} + \frac{π}{2} k = \frac{(3 k - 1) π}{6}$ for integers $k$ . The intervals which comprise the domain are of the form $(x_{k}, x_{k + 1}) = (\frac{(3 k - 1) π}{6}, \frac{(3 k + 2) π}{6})$ as $k$ runs through the integers. Using extended interval notation, we have that the domain is
$⋃_{k = ¯ \infty}^{\infty} (\frac{(3 k - 1) π}{6}, \frac{(3 k + 2) π}{6})$
We can check our answer by substituting in values of $k$ to see that it matches our diagram.
Since the domains of $\sin (x)$ and $\cos (x)$ are all real numbers, the only concern when finding the domain of $f (x) = \frac{\sin (x)}{2 \cos (x) - 1}$ is division by zero so we set the denominator equal to zero and solve. From $2 \cos (x) - 1 = 0$ we get $\cos (x) = \frac{1}{2}$ so that $x = \frac{π}{3} + 2 π k$ or $x = \frac{5 π}{3} + 2 π k$ for integers $k$ . Using set-builder notation, the domain is ${x : x \neq \frac{π}{3} + 2 π k and x \neq \frac{5 π}{3} + 2 π k for integers k}$ , or ${x : x \neq \pm \frac{π}{3}, \pm \frac{5 π}{3}, \pm \frac{7 π}{3}, \pm \frac{11 π}{3}, \dots}$ , so we have
Figure 10.244
Unlike the previous example, we have two different families of points to consider, and we present two ways of dealing with this kind of situation. One way is to generalize what we did in the previous example and use the formulas we found in our domain work to describe the intervals. To that end, we let $a_{k} = \frac{π}{3} + 2 π k = \frac{(6 k + 1) π}{3}$ and $b_{k} = \frac{5 π}{3} + 2 π k = \frac{(6 k + 5) π}{3}$ for integers $k$ . The goal now is to write the domain in terms of the $a$ 's an $b$ 's. We find $a_{0} = \frac{π}{3}$ , $a_{1} = \frac{7 π}{3}$ , $a_{- 1} = - \frac{5 π}{3}$ , $a_{2} = \frac{13 π}{3}$ , $a_{- 2} = - \frac{11 π}{3}$ , $b_{0} = \frac{5 π}{3}$ , $b_{1} = \frac{11 π}{3}$ , $b_{- 1} = - \frac{π}{3}$ , $b_{2} = \frac{17 π}{3}$ and $b_{- 2} = - \frac{7 π}{3}$ . Hence, in terms of the $a$ 's and $b$ 's, our domain is
$\dots (a_{- 2}, b_{- 2}) \cup (b_{- 2}, a_{- 1}) \cup (a_{- 1}, b_{- 1}) \cup (b_{- 1}, a_{0}) \cup (a_{0}, b_{0}) \cup (b_{0}, a_{1}) \cup (a_{1}, b_{1}) \cup \dots$
If we group these intervals in pairs, $(a_{- 2}, b_{- 2}) \cup (b_{- 2}, a_{- 1})$ , $(a_{- 1}, b_{- 1}) \cup (b_{- 1}, a_{0})$ , $open parenthesis a sub 0, b sub 0 close parenthesis union open parenthesis b sub 0, a sub 1 close parenthesis$ and so forth, we see a pattern emerge of the form $open parenthesis a sub k, b sub k close parenthesis union open parenthesis b sub k, a sub k plus 1 close parenthesis$ for integers $k$ so that our domain can be written as
$⋃_{k = ¯ \infty}^{\infty} (a_{k}, b_{k}) \cup (b_{k}, a_{k + 1}) = ⋃_{k = ¯ \infty}^{\infty} (\frac{(6 k + 1) π}{3}, \frac{(6 k + 5) π}{3}) \cup (\frac{(6 k + 5) π}{3}, \frac{(6 k + 7) π}{3})$
A second approach to the problem exploits the periodic nature of $f$ . Since $\cos (x)$ and $\sin (x)$ have period $2 ⁢ π$ , it's not too difficult to show the function $f$ repeats itself every $2 ⁢ π$ units.¹⁵ This means if we can find a formula for the domain on an interval of length $2 ⁢ π$ , we can express the entire domain by translating our answer left and right on the $x$ -axis by adding integer multiples of $2 ⁢ π$ . One such interval that arises from our domain work is $[\frac{π}{3}, \frac{7 π}{3}]$ . The portion of the domain here is $(\frac{π}{3}, \frac{5 π}{3}) \cup (\frac{5 π}{3}, \frac{7 π}{3})$ . Adding integer multiples of $2 ⁢ π$ , we get the family of intervals $(\frac{π}{3} + 2 π k, \frac{5 π}{3} + 2 π k) \cup (\frac{5 π}{3} + 2 π k, \frac{7 π}{3} + 2 π k)$ for integers $k$ . We leave it to the reader to show that getting common denominators leads to our previous answer.
To find the domain of $f (x) = \sqrt{1 - \cot (x)}$ , we first note that, due to the presence of the $\cot (x)$ term, $x \neq π k$ for integers $k$ . Next, we recall that for the square root to be defined, we need $1 minus cot ⁡ open parenthesis x close parenthesis greater than or equal to 0$ . Unlike the inequalities we solved in Example Example 3, we are not restricted here to a given interval. Our strategy is to solve this inequality over $open parenthesis 0, π close parenthesis$ (the same interval which generates a fundamental cycle of cotangent) and then add integer multiples of the period, in this case, $π$ . We let $g (x) = 1 - \cot (x)$ and set about making a sign diagram for $g$ over the interval $open parenthesis 0, π close parenthesis$ to find where $g ⁢ open parenthesis x close parenthesis greater than or equal to 0$ . We note that $g$ is undefined for $x = π k$ for integers $k$ , in particular, at the endpoints of our interval $x equals 0$ and $x = π$ . Next, we look for the zeros of $g$ . Solving $g ⁢ open parenthesis x close parenthesis equals 0$ , we get $cot ⁡ open parenthesis x close parenthesis equals 1$ or $x = \frac{π}{4} + π k$ for integers $k$ and only one of these, $x = \frac{π}{4}$ , lies in $open parenthesis 0, π close parenthesis$ . Choosing the test values $x = \frac{π}{6}$ and $x = \frac{π}{2}$ , we get $g (\frac{π}{6}) = 1 - \sqrt{3}$ , and $g (\frac{π}{2}) = 1$ .
Figure 10.245
We find $g ⁢ open parenthesis x close parenthesis greater than or equal to 0$ on $[\frac{π}{4}, π)$ . Adding multiples of the period we get our solution to consist of the intervals $[\frac{π}{4} + π k, π + π k) = [\frac{(4 k + 1) π}{4}, (k + 1) π)$ . Using extended interval notation, we express our final answer as
$⋃_{k = ¯ \infty}^{\infty} [\frac{(4 k + 1) π}{4}, (k + 1) π)$

We close this section with an example which demonstrates how to solve equations and inequalities involving the inverse trigonometric functions.

Example 5

Solve the following equations and inequalities analytically. Check your answers using a graphing utility.

$\arcsin (2 x) = \frac{π}{3}$
$4 \arccos (x) - 3 π = 0$
$3 arcsec (2 x - 1) + π = 2 π$
$4 \arctan^{2} (x) - 3 π \arctan (x) - π^{2} = 0$
$π squared minus 4 ⁢ arccos squared ⁡ open parenthesis x close parenthesis less than 0$
$4 ⁢ arccot ⁢ open parenthesis 3 ⁢ x close parenthesis greater than π$

To solve $\arcsin (2 x) = \frac{π}{3}$ , we first note that $\frac{π}{3}$ is in the range of the arcsine function (so a solution exists!) Next, we exploit the inverse property of sine and arcsine from Theorem
$\begin{matrix} \arcsin (2 x) & = & \frac{π}{3} \\ \sin (\arcsin (2 x)) & = & \sin (\frac{π}{3}) \\ 2 x & = & \frac{\sqrt{3}}{2} & Since \sin (\arcsin (u)) = u \\ x & = & \frac{\sqrt{3}}{4} \end{matrix}$
Graphing $y equals arcsin ⁡ open parenthesis 2 ⁢ x close parenthesis$ and the horizontal line $y = \frac{π}{3}$ , we see they intersect at $\frac{\sqrt{3}}{4} \approx 0.4430$ .
Our first step in solving $4 \arccos (x) - 3 π = 0$ is to isolate the arccosine. Doing so, we get $\arccos (x) = \frac{3 π}{4}$ . Since $\frac{3 π}{4}$ is in the range of arccosine, we may apply Theorem
$\begin{matrix} \arccos (x) & = & \frac{3 π}{4} \\ \cos (\arccos (x)) & = & \cos (\frac{3 π}{4}) \\ x & = & - \frac{\sqrt{2}}{2} & Since \cos (\arccos (u)) = u \end{matrix}$
The calculator confirms $y = 4 \arccos (x) - 3 π$ crosses $y equals 0$ (the $x$ -axis) at $- \frac{\sqrt{2}}{2} \approx - 0.7071$ .
Figure 10.246
Figure 10.247
$y equals arcsin ⁡ open parenthesis 2 ⁢ x close parenthesis$ and $y = \frac{π}{3}$
$y = 4 \arccos (x) - 3 π$
From $3 arcsec (2 x - 1) + π = 2 π$ , we get $arcsec (2 x - 1) = \frac{π}{3}$ . As we saw in Section, there are two possible ranges for the arcsecant function. Fortunately, both ranges contain $\frac{π}{3}$ . Applying Theorem /, we get
$\begin{matrix} arcsec (2 x - 1) & = & \frac{π}{3} \\ \sec (arcsec (2 x - 1)) & = & \sec (\frac{π}{3}) \\ 2 x - 1 & = & 2 & Since \sec (arcsec (u)) = u \\ x & = & \frac{3}{2} \end{matrix}$
To check using our calculator, we need to graph $y = 3 arcsec (2 x - 1) + π$ . To do so, we make use of the identity $arcsec (u) = \arccos (\frac{1}{u})$ from Theorems and.¹⁶ We see the graph of $y = 3 \arccos (\frac{1}{2 x - 1}) + π$ and the horizontal line $y equals 2 ⁢ π$ intersect at $\frac{3}{2} = 1.5$ .
With the presence of both $\arctan^{2} (x)$ ( $= {(\arctan (x))}^{2}$ ) and $\arctan (x)$ , we substitute $u = \arctan (x)$ . The equation $4 \arctan^{2} (x) - 3 π \arctan (x) - π^{2} = 0$ becomes $4 u^{2} - 3 π u - π^{2} = 0$ . Factoring,¹⁷ we get $(4 u + π) (u - π) = 0$ , so $u = \arctan (x) = - \frac{π}{4}$ or $u = \arctan (x) = π$ . Since $- \frac{π}{4}$ is in the range of arctangent, but $π$ is not, we only get solutions from the first equation. Using Theorem, we get
$\begin{matrix} \arctan (x) & = & - \frac{π}{4} \\ \tan (\arctan (x)) & = & \tan (- \frac{π}{4}) \\ x & = & - 1 & Since \tan (\arctan (u)) = u . \end{matrix}$
The calculator verifies our result.
Figure 10.248
Figure 10.249
$y = 3 arcsec (2 x - 1) + π$ and $y equals 2 ⁢ π$
$y = 4 \arctan^{2} (x) - 3 π \arctan (x) - π^{2}$
Since the inverse trigonometric functions are continuous on their domains, we can solve inequalities featuring these functions using sign diagrams. Since all of the nonzero terms of $π squared minus 4 ⁢ arccos squared ⁡ open parenthesis x close parenthesis less than 0$ are on one side of the inequality, we let $f (x) = π^{2} - 4 \arccos^{2} (x)$ and note the domain of $f$ is limited by the $\arccos (x)$ to $open bracket minus 1, 1 close bracket$ . Next, we find the zeros of $f$ by setting $f (x) = π^{2} - 4 \arccos^{2} (x) = 0$ . We get $\arccos (x) = \pm \frac{π}{2}$ , and since the range of arccosine is $open bracket 0, π close bracket$ , we focus our attention on $\arccos (x) = \frac{π}{2}$ . Using Theorem, we get $x = \cos (\frac{π}{2}) = 0$ as our only zero. Hence, we have two test intervals, $open bracket minus 1, 0 close parenthesis$ and $open parenthesis 0, 1 close bracket$ . Choosing test values $x = \pm 1$ , we get $f (- 1) = - 3 π^{2} < 0$ and $f ⁢ open parenthesis 1 close parenthesis equals π squared greater than 0$ . Since we are looking for where $f (x) = π^{2} - 4 \arccos^{2} (x) < 0$ , our answer is $open bracket minus 1, 0 close parenthesis$ . The calculator confirms that for these values of $x$ , the graph of $y = π^{2} - 4 \arccos^{2} (x)$ is below $y equals 0$ (the $x$ -axis.)
Figure 10.250
Figure 10.251
$y = π^{2} - 4 \arccos^{2} (x)$
To begin, we rewrite $4 ⁢ arccot ⁢ open parenthesis 3 ⁢ x close parenthesis greater than π$ as $4 ⁢ arccot ⁢ open parenthesis 3 ⁢ x close parenthesis minus π greater than 0$ . We let $f (x) = 4 arccot (3 x) - π$ , and note the domain of $f$ is all real numbers, $open parenthesis minus ∞, ∞ close parenthesis$ . To find the zeros of $f$ , we set $f (x) = 4 arccot (3 x) - π = 0$ and solve. We get $arccot (3 x) = \frac{π}{4}$ , and since $\frac{π}{4}$ is in the range of arccotangent, we may apply Theorem and solve
$\begin{matrix} arccot (3 x) & = & \frac{π}{4} \\ \cot (arccot (3 x)) & = & \cot (\frac{π}{4}) \\ 3 x & = & 1 & Since \cot (arccot (u)) = u . \\ x & = & \frac{1}{3} \end{matrix}$
Next, we make a sign diagram for $f$ . Since the domain of $f$ is all real numbers, and there is only one zero of $f$ , $x = \frac{1}{3}$ , we have two test intervals, $(- \infty, \frac{1}{3})$ and $(\frac{1}{3}, \infty)$ . Ideally, we wish to find test values $x$ in these intervals so that $arccot ⁢ open parenthesis 4 ⁢ x close parenthesis$ corresponds to one of our oft-used `common' angles. After a bit of computation,¹⁸ we choose $x equals 0$ for $x < \frac{1}{3}$ and for $x > \frac{1}{3}$ , we choose $x = \frac{\sqrt{3}}{3}$ . We find $f ⁢ open parenthesis 0 close parenthesis equals π greater than 0$ and $f (\frac{\sqrt{3}}{3}) = - \frac{π}{3} < 0$ . Since we are looking for where $f (x) = 4 arccot (3 x) - π > 0$ , we get our answer $(- \infty, \frac{1}{3})$ . To check graphically, we use the technique in number of Example in Section to graph $y equals 4 ⁢ arccot ⁢ open parenthesis 3 ⁢ x close parenthesis$ and we see it is above the horizontal line $y = π$ on $(- \infty, \frac{1}{3}) = (- \infty, 0 . \bar{3})$ .
Figure 10.252
Figure 10.253
$y equals 4 ⁢ arccot ⁢ open parenthesis 3 ⁢ x close parenthesis$ and $y = π$

Exercises

In Exercises -, find all of the exact solutions of the equation and then list those solutions which are in the interval $open bracket 0, 2 ⁢ π close parenthesis$ .

$sin ⁡ open parenthesis 5 ⁢ x close parenthesis equals 0$
$\cos (3 x) = \frac{1}{2}$
$\sin (- 2 x) = \frac{\sqrt{3}}{2}$
$tan ⁡ open parenthesis 6 ⁢ x close parenthesis equals 1$
$\csc (4 x) = - 1$
$\sec (3 x) = \sqrt{2}$
$\cot (2 x) = - \frac{\sqrt{3}}{3}$
$cos ⁡ open parenthesis 9 ⁢ x close parenthesis equals 9$
$\sin (\frac{x}{3}) = \frac{\sqrt{2}}{2}$
$\cos (x + \frac{5 π}{6}) = 0$
$\sin (2 x - \frac{π}{3}) = - \frac{1}{2}$
$2 \cos (x + \frac{7 π}{4}) = \sqrt{3}$
$csc ⁡ open parenthesis x close parenthesis equals 0$
$\tan (2 x - π) = 1$
$tan squared ⁡ open parenthesis x close parenthesis equals 3$
$\sec^{2} (x) = \frac{4}{3}$
$\cos^{2} (x) = \frac{1}{2}$
$\sin^{2} (x) = \frac{3}{4}$
$\sin (x) = \cos (x)$
$sin ⁡ open parenthesis 2 ⁢ x close parenthesis equals sin ⁡ open parenthesis x close parenthesis$
$sin ⁡ open parenthesis 2 ⁢ x close parenthesis equals cos ⁡ open parenthesis x close parenthesis$
$cos ⁡ open parenthesis 2 ⁢ x close parenthesis equals sin ⁡ open parenthesis x close parenthesis$
$cos ⁡ open parenthesis 2 ⁢ x close parenthesis equals cos ⁡ open parenthesis x close parenthesis$
$\cos (2 x) = 2 - 5 \cos (x)$
$3 \cos (2 x) + \cos (x) + 2 = 0$
$\cos (2 x) = 5 \sin (x) - 2$
$3 \cos (2 x) = \sin (x) + 2$
$2 \sec^{2} (x) = 3 - \tan (x)$
$\tan^{2} (x) = 1 - \sec (x)$
$\cot^{2} (x) = 3 \csc (x) - 3$
$sec ⁡ open parenthesis x close parenthesis equals 2 ⁢ csc ⁡ open parenthesis x close parenthesis$
$\cos (x) \csc (x) \cot (x) = 6 - \cot^{2} (x)$
$sin ⁡ open parenthesis 2 ⁢ x close parenthesis equals tan ⁡ open parenthesis x close parenthesis$
$\cot^{4} (x) = 4 \csc^{2} (x) - 7$
$\cos (2 x) + \csc^{2} (x) = 0$
$tan cubed ⁡ open parenthesis x close parenthesis equals 3 ⁢ tan ⁡ open parenthesis x close parenthesis$
$\tan^{2} (x) = \frac{3}{2} \sec (x)$
$\cos^{3} (x) = - \cos (x)$
$\tan (2 x) - 2 \cos (x) = 0$
$\csc^{3} (x) + \csc^{2} (x) = 4 \csc (x) + 4$
$2 \tan (x) = 1 - \tan^{2} (x)$
$\tan (x) = \sec (x)$
$\sin (6 x) \cos (x) = - \cos (6 x) \sin (x)$
$sin ⁡ open parenthesis 3 ⁢ x close parenthesis ⁢ cos ⁡ open parenthesis x close parenthesis equals cos ⁡ open parenthesis 3 ⁢ x close parenthesis ⁢ sin ⁡ open parenthesis x close parenthesis$
$\cos (2 x) \cos (x) + \sin (2 x) \sin (x) = 1$
$\cos (5 x) \cos (3 x) - \sin (5 x) \sin (3 x) = \frac{\sqrt{3}}{2}$
$\sin (x) + \cos (x) = 1$
$\sin (x) + \sqrt{3} \cos (x) = 1$
$\sqrt{2} \cos (x) - \sqrt{2} \sin (x) = 1$
$\sqrt{3} \sin (2 x) + \cos (2 x) = 1$
$\cos (2 x) - \sqrt{3} \sin (2 x) = \sqrt{2}$
$3 \sqrt{3} \sin (3 x) - 3 \cos (3 x) = 3 \sqrt{3}$
$cos ⁡ open parenthesis 3 ⁢ x close parenthesis equals cos ⁡ open parenthesis 5 ⁢ x close parenthesis$
$cos ⁡ open parenthesis 4 ⁢ x close parenthesis equals cos ⁡ open parenthesis 2 ⁢ x close parenthesis$
$sin ⁡ open parenthesis 5 ⁢ x close parenthesis equals sin ⁡ open parenthesis 3 ⁢ x close parenthesis$
$\cos (5 x) = - \cos (2 x)$
$\sin (6 x) + \sin (x) = 0$
$\tan (x) = \cos (x)$
$arccos ⁡ open parenthesis 2 ⁢ x close parenthesis equals π$
$π - 2 \arcsin (x) = 2 π$
$4 \arctan (3 x - 1) - π = 0$
$6 arccot (2 x) - 5 π = 0$
$4 arcsec (\frac{x}{2}) = π$
$12 arccsc (\frac{x}{3}) = 2 π$
$9 \arcsin^{2} (x) - π^{2} = 0$
$9 \arccos^{2} (x) - π^{2} = 0$
$8 {arccot}^{2} (x) + 3 π^{2} = 10 π arccot (x)$
$6 \arctan {(x)}^{2} = π \arctan (x) + π^{2}$
$sin ⁡ open parenthesis x close parenthesis less than or equal to 0$
$tan ⁡ open parenthesis x close parenthesis greater than or equal to the square root of 3$
$sec squared ⁡ open parenthesis x close parenthesis less than or equal to 4$
$\cos^{2} (x) > \frac{1}{2}$
$cos ⁡ open parenthesis 2 ⁢ x close parenthesis less than or equal to 0$
$\sin (x + \frac{π}{3}) > \frac{1}{2}$
$\cot^{2} (x) \geq \frac{1}{3}$
$2 ⁢ cos ⁡ open parenthesis x close parenthesis greater than or equal to 1$
$sin ⁡ open parenthesis 5 ⁢ x close parenthesis greater than or equal to 5$
$cos ⁡ open parenthesis 3 ⁢ x close parenthesis less than or equal to 1$
$sec ⁡ open parenthesis x close parenthesis less than or equal to the square root of 2$
$cot ⁡ open parenthesis x close parenthesis less than or equal to 4$
$\cos (x) > \frac{\sqrt{3}}{2}$
$\sin (x) > \frac{1}{3}$
$sec ⁡ open parenthesis x close parenthesis less than or equal to 2$
$\sin^{2} (x) < \frac{3}{4}$
$cot ⁡ open parenthesis x close parenthesis greater than or equal to minus 1$
$\cos (x) \geq \sin (x)$
$csc ⁡ open parenthesis x close parenthesis greater than 1$
$\cos (x) \leq \frac{5}{3}$
$cot ⁡ open parenthesis x close parenthesis greater than or equal to 5$
$tan squared ⁡ open parenthesis x close parenthesis greater than or equal to 1$
$sin ⁡ open parenthesis 2 ⁢ x close parenthesis greater than or equal to sin ⁡ open parenthesis x close parenthesis$
$cos ⁡ open parenthesis 2 ⁢ x close parenthesis less than or equal to sin ⁡ open parenthesis x close parenthesis$
$arcsin ⁡ open parenthesis 2 ⁢ x close parenthesis greater than 0$
$3 ⁢ arccos ⁡ open parenthesis x close parenthesis less than or equal to π$
$6 ⁢ arccot ⁢ open parenthesis 7 ⁢ x close parenthesis greater than or equal to π$
$π greater than 2 ⁢ arctan ⁡ open parenthesis x close parenthesis$
$2 arcsin open parenthesis x close parenthesis squared greater than π arcsin open parenthesis x close parenthesis$
$12 arccos open parenthesis x close parenthesis squared plus 2 π squared greater than 11 π arccos open parenthesis x close parenthesis$
$f (x) = \frac{1}{\cos (x) - 1}$
$f (x) = \frac{\cos (x)}{\sin (x) + 1}$
$f (x) = \sqrt{\tan^{2} (x) - 1}$
$f (x) = \sqrt{2 - \sec (x)}$
$f ⁢ open parenthesis x close parenthesis equals csc ⁡ open parenthesis 2 ⁢ x close parenthesis$
$f (x) = \frac{\sin (x)}{2 + \cos (x)}$
$f (x) = 3 \csc (x) + 4 \sec (x)$
$f (x) = \ln (| \cos (x) |)$
$f (x) = \arcsin (\tan (x))$
With the help of your classmates, determine the number of solutions to $\sin (x) = \frac{1}{2}$ in $open bracket 0, 2 ⁢ π close parenthesis$ . Then find the number of solutions to $\sin (2 x) = \frac{1}{2}$ , $\sin (3 x) = \frac{1}{2}$ and $\sin (4 x) = \frac{1}{2}$ in $open bracket 0, 2 ⁢ π close parenthesis$ . A pattern should emerge. Explain how this pattern would help you solve equations like $\sin (11 x) = \frac{1}{2}$ . Now consider $\sin (\frac{x}{2}) = \frac{1}{2}$ , $\sin (\frac{3 x}{2}) = \frac{1}{2}$ and $\sin (\frac{5 x}{2}) = \frac{1}{2}$ . What do you find? Replace $\frac{1}{2}$ with $minus 1$ and repeat the whole exploration.

In Exercises -, solve the equation, giving the exact solutions which lie in $open bracket 0, 2 ⁢ π close parenthesis$

In Exercises -, solve the equation.

In Exercises -, solve the inequality. Express the exact answer in interval notation, restricting your attention to $0 less than or equal to x less than or equal to 2 ⁢ π$ .

In Exercises -, solve the inequality. Express the exact answer in interval notation, restricting your attention to $- π \leq x \leq π$ .

In Exercises -, solve the inequality. Express the exact answer in interval notation, restricting your attention to $minus 2 ⁢ π less than or equal to x less than or equal to 2 ⁢ π$ .

In Exercises -, solve the given inequality.

In Exercises -, express the domain of the function using the extended interval notation. (See page in Section for details.)

Answers

$x = \frac{π k}{5}; x = 0, \frac{π}{5}, \frac{2 π}{5}, \frac{3 π}{5}, \frac{4 π}{5}, π, \frac{6 π}{5}, \frac{7 π}{5}, \frac{8 π}{5}, \frac{9 π}{5}$
$x = \frac{π}{9} + \frac{2 π k}{3}$ or $x = \frac{5 π}{9} + \frac{2 π k}{3}; x = \frac{π}{9}, \frac{5 π}{9}, \frac{7 π}{9}, \frac{11 π}{9}, \frac{13 π}{9}, \frac{17 π}{9}$
$x = \frac{2 π}{3} + π k$ or $x = \frac{5 π}{6} + π k; x = \frac{2 π}{3}, \frac{5 π}{6}, \frac{5 π}{3}, \frac{11 π}{6}$
$x = \frac{π}{24} + \frac{π k}{6}; x = \frac{π}{24}, \frac{5 π}{24}, \frac{3 π}{8}, \frac{13 π}{24}, \frac{17 π}{24}, \frac{7 π}{8}, \frac{25 π}{24}, \frac{29 π}{24}, \frac{11 π}{8}, \frac{37 π}{24}, \frac{41 π}{24}, \frac{15 π}{8}$
$x = \frac{3 π}{8} + \frac{π k}{2}; x = \frac{3 π}{8}, \frac{7 π}{8}, \frac{11 π}{8}, \frac{15 π}{8}$
$x = \frac{π}{12} + \frac{2 π k}{3}$ or $x = \frac{7 π}{12} + \frac{2 π k}{3}; x = \frac{π}{12}, \frac{7 π}{12}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{17 π}{12}, \frac{23 π}{12}$
$x = \frac{π}{3} + \frac{π k}{2}; x = \frac{π}{3}, \frac{5 π}{6}, \frac{4 π}{3}, \frac{11 π}{6}$
No solution
$x = \frac{3 π}{4} + 6 π k$ or $x = \frac{9 π}{4} + 6 π k; x = \frac{3 π}{4}$
$x = - \frac{π}{3} + π k; x = \frac{2 π}{3}, \frac{5 π}{3}$
$x = \frac{3 π}{4} + π k$ or $x = \frac{13 π}{12} + π k; x = \frac{π}{12}, \frac{3 π}{4}, \frac{13 π}{12}, \frac{7 π}{4}$
$x = - \frac{19 π}{12} + 2 π k$ or $x = \frac{π}{12} + 2 π k; x = \frac{π}{12}, \frac{5 π}{12}$
No solution
$x = \frac{5 π}{8} + \frac{π k}{2}; x = \frac{π}{8}, \frac{5 π}{8}, \frac{9 π}{8}, \frac{13 π}{8}$
$x = \frac{π}{3} + π k$ or $x = \frac{2 π}{3} + π k; x = \frac{π}{3}, \frac{2 π}{3}, \frac{4 π}{3}, \frac{5 π}{3}$
$x = \frac{π}{6} + π k$ or $x = \frac{5 π}{6} + π k; x = \frac{π}{6}, \frac{5 π}{6}, \frac{7 π}{6}, \frac{11 π}{6}$
$x = \frac{π}{4} + \frac{π k}{2}; x = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$
$x = \frac{π}{3} + π k$ or $x = \frac{2 π}{3} + π k; x = \frac{π}{3}, \frac{2 π}{3}, \frac{4 π}{3}, \frac{5 π}{3}$
$x = \frac{π}{4}, \frac{5 π}{4}$
$x = 0, \frac{π}{3}, π, \frac{5 π}{3}$
$x = \frac{π}{6}, \frac{π}{2}, \frac{5 π}{6}, \frac{3 π}{2}$
$x = \frac{π}{6}, \frac{5 π}{6}, \frac{3 π}{2}$
$x = 0, \frac{2 π}{3}, \frac{4 π}{3}$
$x = \frac{π}{3}, \frac{5 π}{3}$
$x = \frac{2 π}{3}, \frac{4 π}{3}, \arccos (\frac{1}{3}), 2 π - \arccos (\frac{1}{3})$
$x = \frac{π}{6}, \frac{5 π}{6}$
$x = \frac{7 π}{6}, \frac{11 π}{6}, \arcsin (\frac{1}{3}), π - \arcsin (\frac{1}{3})$
$x = \frac{3 π}{4}, \frac{7 π}{4}, \arctan (\frac{1}{2}), π + \arctan (\frac{1}{2})$
$x = 0, \frac{2 π}{3}, \frac{4 π}{3}$
$x = \frac{π}{6}, \frac{5 π}{6}, \frac{π}{2}$
$x = \arctan (2), π + \arctan (2)$
$x = \frac{π}{6}, \frac{7 π}{6}, \frac{5 π}{6}, \frac{11 π}{6}$
$x = 0, π, \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$
$x = \frac{π}{6}, \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{6}, \frac{7 π}{6}, \frac{5 π}{4}, \frac{7 π}{4}, \frac{11 π}{6}$
$x = \frac{π}{2}, \frac{3 π}{2}$
$x = 0, \frac{π}{3}, \frac{2 π}{3}, π, \frac{4 π}{3}, \frac{5 π}{3}$
$x = \frac{π}{3}, \frac{5 π}{3}$
$x = \frac{π}{2}, \frac{3 π}{2}$
$x = \frac{π}{6}, \frac{π}{2}, \frac{5 π}{6}, \frac{3 π}{2}$
$x = \frac{π}{6}, \frac{5 π}{6}, \frac{7 π}{6}, \frac{3 π}{2}, \frac{11 π}{6}$
$x = \frac{π}{8}, \frac{5 π}{8}, \frac{9 π}{8}, \frac{13 π}{8}$
No solution
$x = 0, \frac{π}{7}, \frac{2 π}{7}, \frac{3 π}{7}, \frac{4 π}{7}, \frac{5 π}{7}, \frac{6 π}{7}, π, \frac{8 π}{7}, \frac{9 π}{7}, \frac{10 π}{7}, \frac{11 π}{7}, \frac{12 π}{7}, \frac{13 π}{7}$
$x = 0, \frac{π}{2}, π, \frac{3 π}{2}$
$x equals 0$
$x = \frac{π}{48}, \frac{11 π}{48}, \frac{13 π}{48}, \frac{23 π}{48}, \frac{25 π}{48}, \frac{35 π}{48}, \frac{37 π}{48}, \frac{47 π}{48}, \frac{49 π}{48}, \frac{59 π}{48}, \frac{61 π}{48}, \frac{71 π}{48}, \frac{73 π}{48}, \frac{83 π}{48}, \frac{85 π}{48}, \frac{95 π}{48}$
$x = 0, \frac{π}{2}$
$x = \frac{π}{2}, \frac{11 π}{6}$
$x = \frac{π}{12}, \frac{17 π}{12}$
$x = 0, π, \frac{π}{3}, \frac{4 π}{3}$
$x = \frac{17 π}{24}, \frac{41 π}{24}, \frac{23 π}{24}, \frac{47 π}{24}$
$x = \frac{π}{6}, \frac{5 π}{18}, \frac{5 π}{6}, \frac{17 π}{18}, \frac{3 π}{2}, \frac{29 π}{18}$
$x = 0, \frac{π}{4}, \frac{π}{2}, \frac{3 π}{4}, π, \frac{5 π}{4}, \frac{3 π}{2}, \frac{7 π}{4}$
$x = 0, \frac{π}{3}, \frac{2 π}{3}, π, \frac{4 π}{3}, \frac{5 π}{3}$
$x = 0, \frac{π}{8}, \frac{3 π}{8}, \frac{5 π}{8}, \frac{7 π}{8}, π, \frac{9 π}{8}, \frac{11 π}{8}, \frac{13 π}{8}, \frac{15 π}{8}$
$x = \frac{π}{7}, \frac{π}{3}, \frac{3 π}{7}, \frac{5 π}{7}, π, \frac{9 π}{7}, \frac{11 π}{7}, \frac{5 π}{3}, \frac{13 π}{7}$
$x = 0, \frac{2 π}{7}, \frac{4 π}{7}, \frac{6 π}{7}, \frac{8 π}{7}, \frac{10 π}{7}, \frac{12 π}{7}, \frac{π}{5}, \frac{3 π}{5}, π, \frac{7 π}{5}, \frac{9 π}{5}$
$x = \arcsin (\frac{- 1 + \sqrt{5}}{2}) \approx 0.6662, π - \arcsin (\frac{- 1 + \sqrt{5}}{2}) \approx 2.4754$
$x = - \frac{1}{2}$
$x = - 1$
$x = \frac{2}{3}$
$x = - \frac{\sqrt{3}}{2}$
$x = 2 \sqrt{2}$
$x equals 6$
$x = \pm \frac{\sqrt{3}}{2}$
$x = \frac{1}{2}$
$x = - 1, 0$
$x = - \sqrt{3}$
$open bracket π, 2 ⁢ π close bracket$
$[\frac{π}{3}, \frac{π}{2}) \cup [\frac{4 π}{3}, \frac{3 π}{2})$
$[0, \frac{π}{3}] \cup [\frac{2 π}{3}, \frac{4 π}{3}] \cup [\frac{5 π}{3}, 2 π]$
$[0, \frac{π}{4}) \cup (\frac{3 π}{4}, \frac{5 π}{4}) \cup (\frac{7 π}{4}, 2 π]$
$[\frac{π}{4}, \frac{3 π}{4}] \cup [\frac{5 π}{4}, \frac{7 π}{4}]$
$[0, \frac{π}{2}) \cup (\frac{11 π}{6}, 2 π]$
$(0, \frac{π}{3}] \cup [\frac{2 π}{3}, π) \cup (π, \frac{4 π}{3}] \cup [\frac{5 π}{3}, 2 π)$
$[0, \frac{π}{3}] \cup [\frac{5 π}{3}, 2 π]$
No solution
$open bracket 0, 2 ⁢ π close bracket$
$[0, \frac{π}{4}] \cup (\frac{π}{2}, \frac{3 π}{2}) \cup [\frac{7 π}{4}, 2 π]$
$open bracket arccot ⁢ open parenthesis 4 close parenthesis, π close parenthesis union open bracket π plus arccot ⁢ open parenthesis 4 close parenthesis, 2 ⁢ π close parenthesis$
$(- \frac{π}{6}, \frac{π}{6})$
$(\arcsin (\frac{1}{3}), π - \arcsin (\frac{1}{3}))$
$[- π, - \frac{π}{2}) \cup [- \frac{π}{3}, \frac{π}{3}] \cup (\frac{π}{2}, π]$
$(- \frac{2 π}{3}, - \frac{π}{3}) \cup (\frac{π}{3}, \frac{2 π}{3})$
$(- π, - \frac{π}{4}] \cup (0, \frac{3 π}{4}]$
$[- \frac{3 π}{4}, \frac{π}{4}]$
$(- 2 π, - \frac{3 π}{2}) \cup (- \frac{3 π}{2}, - π) \cup (0, \frac{π}{2}) \cup (\frac{π}{2}, π)$
$open bracket minus 2 ⁢ π, 2 ⁢ π close bracket$
$(- 2 π, arccot (5) - 2 π] \cup (- π, arccot (5) - π] \cup (0, arccot (5)] \cup (π, π + arccot (5)]$
$[- \frac{7 π}{4}, - \frac{3 π}{2}) \cup (- \frac{3 π}{2}, - \frac{5 π}{4}] \cup [- \frac{3 π}{4}, - \frac{π}{2}) \cup (- \frac{π}{2}, - \frac{π}{4}] \cup [\frac{π}{4}, \frac{π}{2}) \cup (\frac{π}{2}, \frac{3 π}{4}] \cup [\frac{5 π}{4}, \frac{3 π}{2}) \cup (\frac{3 π}{2}, \frac{7 π}{4}]$
$[- 2 π, - \frac{5 π}{3}] \cup [- π, - \frac{π}{3}] \cup [0, \frac{π}{3}] \cup [π, \frac{5 π}{3}]$
$[- \frac{11 π}{6}, - \frac{7 π}{6}] \cup [\frac{π}{6}, \frac{5 π}{6}] \cup, {- \frac{π}{2}, \frac{3 π}{2}}$
$(0, \frac{1}{2}]$
$[\frac{1}{2}, 1]$
$(- \infty, \frac{\sqrt{3}}{7}]$
$open parenthesis minus ∞, ∞ close parenthesis$
$open bracket minus 1, 0 close parenthesis$
$[- 1, - \frac{1}{2}) \cup (\frac{\sqrt{2}}{2}, 1]$
$⋃_{k = ¯ \infty}^{\infty} (2 k π, (2 k + 2) π)$
$⋃_{k = ¯ \infty}^{\infty} (\frac{(4 k - 1) π}{2}, \frac{(4 k + 3) π}{2})$
$⋃_{k = ¯ \infty}^{\infty} {[\frac{(4 k + 1) π}{4}, \frac{(2 k + 1) π}{2}) \cup (\frac{(2 k + 1) π}{2}, \frac{(4 k + 3) π}{4}]}$
$⋃_{k = ¯ \infty}^{\infty} {[\frac{(6 k - 1) π}{3}, \frac{(6 k + 1) π}{3}] \cup (\frac{(4 k + 1) π}{2}, \frac{(4 k + 3) π}{2})}$
$⋃_{k = ¯ \infty}^{\infty} (\frac{k π}{2}, \frac{(k + 1) π}{2})$
$open parenthesis minus ∞, ∞ close parenthesis$
$⋃_{k = ¯ \infty}^{\infty} (\frac{k π}{2}, \frac{(k + 1) π}{2})$
$⋃_{k = ¯ \infty}^{\infty} (\frac{(2 k - 1) π}{2}, \frac{(2 k + 1) π}{2})$
$⋃_{k = ¯ \infty}^{\infty} [\frac{(4 k - 1) π}{4}, \frac{(4 k + 1) π}{4}]$

See the comments at the beginning of Section for a review of this concept. ↩
Don't forget to divide the $2 ⁢ π ⁢ k$ by $3$ as well! ↩
Do you see why? ↩
The reader is encouraged to see what happens if we had chosen the reciprocal identity $\cot (3 x) = \frac{1}{\tan (3 x)}$ instead. The graph on the calculator appears identical, but what happens when you try to find the intersection points? ↩
Your instructor will let you know if you should abandon the analytic route at this point and use your calculator. But seriously, what fun would that be? ↩
Note that we are not counting the point $open parenthesis 2 ⁢ π, 0 close parenthesis$ in our solution set since $x equals 2 ⁢ π$ is not in the interval $open bracket 0, 2 ⁢ π close parenthesis$ . In the forthcoming solutions, remember that while $x equals 2 ⁢ π$ may be a solution to the equation, it isn't counted among the solutions in $open bracket 0, 2 ⁢ π close parenthesis$ . ↩
As always, experience is the greatest teacher here! ↩
As always, when in doubt, write it out! ↩
We are essentially `undoing' the sum / difference formula for cosine or sine, depending on which form we use, so this problem is actually closely related to the previous one! ↩
See page , Example , page , page , Example and Example for discussion of this technique. ↩
See page for a discussion of the non-standard character known as the interrobang. ↩
We could have chosen any value $\arctan (t)$ where $t greater than 3$ . ↩
…by adding $π$ through the inequality … ↩
See page for details about this notation. ↩
This doesn't necessarily mean the period of $f$ is $2 ⁢ π$ . The tangent function is comprised of $\cos (x)$ and $\sin (x)$ , but its period is half theirs. The reader is invited to investigate the period of $f$ . ↩
Since we are checking for solutions where arcsecant is positive, we know $u = 2 x - 1 \geq 1$ , and so the identity applies in both cases. ↩
It's not as bad as it looks... don't let the $π$ throw you! ↩
Set $3 ⁢ x$ equal to the cotangents of the `common angles' and choose accordingly. ↩

Adapted from Precalculus, 3rd corrected edition, by Carl Stitz and Jeff Zeager (stitz-zeager.com), licensed under CC BY-NC-SA 3.0. Changes were made: reformatted as an accessible XYZ web edition. License: CC-BY-NC-SA-3.0.