<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Skip Navigation

14.3: Solving Trigonometric Equations

Difficulty Level: At Grade Created by: CK-12
Turn In


Here you’ll learn how to solve equations with trigonometric functions.

Review Queue

Solve the following equations.

1. \begin{align*}x^2-7x-18=0\end{align*}

2. \begin{align*}9x^4-16=0\end{align*}

3. \begin{align*}\sin x= \frac{\sqrt{2}}{2},0< x < \frac{\pi}{2}\end{align*}

Using Algebra


Here you'll solve trig equations using algebra.


In the previous concept, we verified trigonometric identities, which are true for every real value of \begin{align*}x\end{align*}. In this concept, we will solve trigonometric equations. An equation is only true for some values of \begin{align*}x\end{align*}.

Example A

Verify that \begin{align*}\csc x-2=0\end{align*} when \begin{align*}x=\frac{5 \pi}{6}\end{align*}.

Solution: Substitute in \begin{align*}x=\frac{5 \pi}{6}\end{align*} to see if the equations holds true.

\begin{align*}\csc \left(\frac{5 \pi}{6}\right)-2&=0 \\ \frac{1}{\sin \left(\frac{5 \pi}{6}\right)}-2&=0 \\ \frac{1}{\frac{1}{2}}-2&=0 \\ 2-2&=0\end{align*}

This is a true statement, so \begin{align*}x=\frac{5 \pi}{6}\end{align*} is a solution to the equation.

Example B

Solve \begin{align*}2 \cos x+1=0\end{align*}.

Solution: To solve this equation, we need to isolate \begin{align*}\cos x\end{align*} and then use inverse to find the values of \begin{align*}x\end{align*} when the equation is valid. You already did this to find the zeros in the graphing concepts earlier in this chapter.

\begin{align*}2 \cos x+1&=0 \\ 2 \cos x&=-1 \\ \cos x&=- \frac{1}{2}\end{align*}

So, when is the \begin{align*}\cos x=- \frac{1}{2}\end{align*}? Between \begin{align*}0 \le x< 2 \pi, x=\frac{2 \pi}{3}\end{align*} and \begin{align*}\frac{4 \pi}{3}\end{align*}. But, the trig functions are periodic, so there are more solutions than just these two. You can write the general solutions as \begin{align*}x=\frac{2 \pi}{3} \pm 2 \pi n\end{align*} and \begin{align*}x=\frac{4 \pi}{3} \pm 2 \pi n\end{align*}, where \begin{align*}n\end{align*} is any integer. You can check your answer graphically by graphing \begin{align*}y=\cos x\end{align*} and \begin{align*}y=- \frac{1}{2}\end{align*} on the same set of axes. Where the two lines intersect are the solutions.

Example C

Solve \begin{align*}5 \tan(x+2)-1=0\end{align*}, where \begin{align*}0 \le x < 2 \pi\end{align*}.

Solution: In this example, we have an interval where we want to find \begin{align*}x\end{align*}. Therefore, at the end of the problem, we will need to add or subtract \begin{align*}\pi\end{align*}, the period of tangent, to find the correct solutions within our interval.

\begin{align*}5 \tan(x+2)-1&=0 \\ 5 \tan(x+2)&=1 \\ \tan(x+2)&=\frac{1}{5}\end{align*}

Using the \begin{align*}\tan^{-1}\end{align*} button on your calculator, we get that \begin{align*}\tan^{-1} \left(\frac{1}{5}\right)=0.1974\end{align*}. Therefore, we have:

\begin{align*}x+2&=0.1974 \\ x&=-1.8026\end{align*}

This answer is not within our interval. To find the solutions in the interval, add \begin{align*}\pi\end{align*} a couple of times until we have found all of the solutions in \begin{align*}[0, 2 \pi]\end{align*}.

\begin{align*}x&=-1.8026+ \pi=1.3390 \\ &=1.3390+ \pi=4.4806\end{align*}

The two solutions are \begin{align*}x = 1.3390\end{align*} and 4.4806.

Guided Practice

1. Determine if \begin{align*}x=\frac{\pi}{3}\end{align*} is a solution for \begin{align*}2 \sin x=\sqrt{3}\end{align*}.

Solve the following trig equations in the interval \begin{align*}0 \le x< 2 \pi\end{align*}.

2. \begin{align*}3 \cos^2x-5=0\end{align*}

3. \begin{align*}3 \sec(x-1)+2=0\end{align*}


1. \begin{align*}2 \sin \frac{\pi}{3}= \sqrt{3} \rightarrow 2 \cdot \frac{\sqrt{3}}{2}=\sqrt{3}\end{align*} Yes, \begin{align*}x=\frac{\pi}{3}\end{align*} is a solution.

2. Isolate the \begin{align*}\cos^2x\end{align*} and then take the square root of both sides. Don’t forget about the \begin{align*}\pm\end{align*}!

\begin{align*}9 \cos^2x-5&=0 \\ 9 \cos^2x&=5 \\ \cos^2x&=\frac{5}{9} \\ \cos x&=\pm \frac{\sqrt{5}}{3}\end{align*}

The \begin{align*}\cos x=\frac{\sqrt{5}}{3}\end{align*} at \begin{align*}x=0.243\end{align*} rad (use your graphing calculator). To find the other value where cosine is positive, subtract 0.243 from \begin{align*}2 \pi\end{align*}, \begin{align*}x=2 \pi -0.243=6.037\end{align*} rad.

The \begin{align*}\cos x=- \frac{\sqrt{5}}{3}\end{align*} at \begin{align*}x=2.412\end{align*} rad, which is in the \begin{align*}2^{nd}\end{align*} quadrant. To find the other value where cosine is negative (the \begin{align*}3^{rd}\end{align*} quadrant), use the reference angle, 0.243, and add it to \begin{align*}\pi\end{align*}. \begin{align*}x= \pi+0.243=3.383\end{align*} rad.

3. Here, we will find the solution within the given range, \begin{align*}0 \le x< 2 \pi\end{align*}.

\begin{align*}3 \sec(x-1)+2&=0 \\ 3 \sec(x-1)&=-2 \\ \sec(x-1)&=- \frac{2}{3} \\ \cos(x-1)&=- \frac{3}{2}\end{align*}

At this point, we can stop. The range of the cosine function is from 1 to -1. \begin{align*}- \frac{3}{2}\end{align*} is outside of this range, so there is no solution to this equation.

Problem Set

Determine if the following values for \begin{align*}x\end{align*}. are solutions to the equation \begin{align*}5+6 \csc x=17\end{align*}.

  1. \begin{align*}x=- \frac{7 \pi}{6}\end{align*}
  2. \begin{align*}x=\frac{11 \pi}{6}\end{align*}
  3. \begin{align*}x=\frac{5 \pi}{6}\end{align*}

Solve the following trigonometric equations. If no solutions exist, write no solution.

  1. \begin{align*}1- \cos x=0\end{align*}
  2. \begin{align*}3 \tan x - \sqrt{3}=0\end{align*}
  3. \begin{align*}4 \cos x=2 \cos x+1\end{align*}
  4. \begin{align*}5 \sin x-2=2 \sin x+4\end{align*}
  5. \begin{align*}\sec x-4=- \sec x\end{align*}
  6. \begin{align*}\tan^2(x-2)=3\end{align*}

Sole the following trigonometric equations within the interval \begin{align*}0 \le x < 2 \pi\end{align*}. If no solutions exist, write no solution.

  1. \begin{align*}\cos x=\sin x\end{align*}
  2. \begin{align*}- \sqrt{3} \csc x=2\end{align*}
  3. \begin{align*}6 \sin(x-2)=14\end{align*}
  4. \begin{align*}7 \cos x -4=1\end{align*}
  5. \begin{align*}5+4 \cot^2x=17\end{align*}
  6. \begin{align*}2 \sin^2x-7=-6\end{align*}

By Using Quadratic Techniques


Here you'll solve trig equations by factoring and the Quadratic Formula.


Another way to solve a trig equation is to use factoring or the quadratic formula. Let’s look at a couple of examples.

Example A

Solve \begin{align*}\sin^2x-3 \sin x+2=0\end{align*}.

Solution: This sine equation looks a lot like the quadratic \begin{align*}x^2-3x+2=0\end{align*} which factors to be \begin{align*}(x-2)(x-1)=0\end{align*} and the solutions are \begin{align*}x = 2\end{align*} and 1. We can factor the trig equation in the exact same manner. Instead of just \begin{align*}x\end{align*}, we will have \begin{align*}\sin x\end{align*} in the factors.

\begin{align*}\sin^2x-3 \sin x+2&=0 \\ (\sin x-2)(\sin x-1)&=0 \\ \sin x=2 \ and \ \sin x&=1\end{align*}

There is no solution for \begin{align*}\sin x=2\end{align*} and \begin{align*}\sin x=1\end{align*} when \begin{align*}x= \frac{\pi}{2} \pm 2 \pi n\end{align*}.

Example B

Solve \begin{align*}1- \sin x=\sqrt{3} \cos x\end{align*} in the interval \begin{align*}0 \le x < 2 \pi\end{align*}.

Solution: To solve this equation, use the Pythagorean Identity \begin{align*}\sin^2x+\cos^2x=1\end{align*}. Solve for either cosine and substitute into the equation. \begin{align*}\cos x=\sqrt{1- \sin^2 x}\end{align*}

\begin{align*}1- \sin x&=\sqrt{3} \cdot \sqrt{1- \sin^2x} \\ (1- \sin x)^2&=\sqrt{3-3 \sin^2x}^2 \\ 1-2 \sin x+\sin^2x&=3-3 \sin^2x \\ 4 \sin^2x-2 \sin x-2&=0 \\ 2 \sin^2x-\sin x-1&=0 \\ (2 \sin x+1)(\sin x-1)&=0\end{align*}

Solving each factor for \begin{align*}x\end{align*}, we get \begin{align*}\sin x=- \frac{1}{2} \rightarrow x=\frac{7 \pi}{6}\end{align*} and \begin{align*}\frac{11 \pi}{6}\end{align*} and \begin{align*}\sin x=1 \rightarrow x=\frac{\pi}{2}\end{align*}.

Example C

Solve \begin{align*}\tan^2x-5 \tan x-9=0\end{align*} in the interval \begin{align*}0 \le x < \pi\end{align*}.

Solution: This equation is not factorable so you have to use the Quadratic Formula.

\begin{align*}\tan x&=\frac{5 \pm \sqrt{\left(-5\right)^2-4 \left(1\right) \left(-9\right)}}{2} \\ &=\frac{5 \pm \sqrt{61}}{2} \\ & \approx 6.41 \ and \ -1.41\end{align*}

\begin{align*}x \approx \tan^{-1} 6.41 \approx 1.416 \ \text{rad}\end{align*} and \begin{align*}x \approx \tan^{-1}-1.41 \approx -0.954 \ \text{rad}\end{align*}

The first answer is within the range, but the second is not. To adjust -0.954 to be within the range, we need to find the answer in the second quadrant, \begin{align*}\pi - 0.954=2.186 \ \text{rad}\end{align*}.

Guided Practice

Solve the following trig equations using any method in the interval \begin{align*}0 \le x < 2 \pi\end{align*}.

1. \begin{align*}\sin^2x \cos x=\cos x\end{align*}

2. \begin{align*}\sin^2x=2 \sin(-x)+1\end{align*}

3. \begin{align*}4 \cos^2x-2 \cos x-1=0\end{align*}


1. Put everything onto one side of the equation and factor out a cosine.

\begin{align*}\sin^2x \cos x- \cos x&=0 \\ \cos x(\sin^2x-1)&=0 \\ \cos x(\sin x -1)(\sin x+1)&=0\end{align*}

\begin{align*}\cos x&=0 \qquad \qquad \ \ \sin x=1 \qquad \ \ \sin x=-1 \\ x&=\frac{\pi}{2} \ and \ \frac{3 \pi}{2} \qquad \ x=\frac{\pi}{2} \qquad \qquad x=\frac{3 \pi}{2}\end{align*}

2. Recall that \begin{align*}\sin(-x)=- \sin x\end{align*} from the Negative Angle Identities.

\begin{align*}\sin^2x&=2 \sin(-x)+1 \\ \sin^2x&=-2 \sin x+1 \\ \sin^2x+2 \sin x+1&=0 \\ (\sin x+1)^2&=0 \\ \sin x&=-1 \\ x&=\frac{3 \pi}{2}\end{align*}

3. This quadratic is not factorable, so use the quadratic formula.

\begin{align*}\cos x&=\frac{2 \pm \sqrt{2^2 -4 \left(4\right) \left(-1\right)}}{2 \left(4\right)} \\ &=\frac{2 \pm \sqrt{20}}{8} \\ &=\frac{1 \pm 2 \sqrt{5}}{4}\end{align*}

\begin{align*}x& \approx \cos^{-1} \left(\frac{1+ \sqrt{5}}{4}\right) && x \approx \cos^{-1} \left(\frac{1- \sqrt{5}}{4}\right) \\ & \approx \cos^{-1} 0.8090 \qquad and && \ \ \approx \cos^{-1}-0.3090 \\ & \approx 0.6283 && \ \ \approx 1.8850 \ (\text{reference angle is} \ \pi-1.8850 \approx 1.2570)\end{align*}

The other solutions in the range are \begin{align*}x \approx 2 \pi - 0.6283 \approx 5.6549\end{align*} and \begin{align*}x \approx \pi + 1.2570 \approx 4.3982\end{align*}.

Problem Set

Solve the following trig equations using any method. Find all solutions in the interval \begin{align*}0 \le x < 2 \pi\end{align*}. Round any decimal answers to 4 decimal places.

  1. \begin{align*}2 \cos^2x-\sin x -1=0\end{align*}
  2. \begin{align*}4 \sin^2x+5 \sin x+1=0\end{align*}
  3. \begin{align*}3 \tan^2x- \tan x=0\end{align*}
  4. \begin{align*}2 \cos^2x+\cos(-x)-1=0\end{align*}
  5. \begin{align*}1- \sin x=\sqrt{2} \cos x\end{align*}
  6. \begin{align*}\sqrt{\sin x}=2 \sin x-1\end{align*}
  7. \begin{align*}\sin^3x-\sin x=0\end{align*}
  8. \begin{align*}\tan^2x-8 \tan x-7=0\end{align*}
  9. \begin{align*}5 \cos^2x+3 \cos x-2=0\end{align*}
  10. \begin{align*}\sin x- \sin x \cos^2x=1\end{align*}
  11. \begin{align*}\cos^2x-3 \cos x-2=0\end{align*}
  12. \begin{align*}\sin^2x \cos x=4 \cos x\end{align*}
  13. \begin{align*}\cos x \csc^2x+2 \cos x=6 \cos x\end{align*}

Using your graphing calculator, graph the following equations and determine the points of intersection in the interval \begin{align*}0 \le x < 2 \pi\end{align*}.

  1. .
\begin{align*}y&=\sin^2x \\ y&=2 \sin x-1\end{align*}
  1. .
\begin{align*}y&=4 \cos x-3 \\ y&=-2 \tan x\end{align*}

Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes
Please to create your own Highlights / Notes
Show More

Image Attributions

Show Hide Details
Date Created:
Apr 23, 2013
Last Modified:
Jul 06, 2016
Save or share your relevant files like activites, homework and worksheet.
To add resources, you must be the owner of the section. Click Customize to make your own copy.
Please wait...
Please wait...
Image Detail
Sizes: Medium | Original