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14.3: Solving Trigonometric Equations

Difficulty Level: At Grade Created by: CK-12
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Here you’ll learn how to solve equations with trigonometric functions.

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Solve the following equations.

1. \begin{align*}x^2-7x-18=0\end{align*}x27x18=0

2. \begin{align*}9x^4-16=0\end{align*}9x416=0

3. \begin{align*}\sin x= \frac{\sqrt{2}}{2},0< x < \frac{\pi}{2}\end{align*}sinx=22,0<x<π2

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*}x. In this concept, we will solve trigonometric equations. An equation is only true for some values of \begin{align*}x\end{align*}x.

Example A

Verify that \begin{align*}\csc x-2=0\end{align*}cscx2=0 when \begin{align*}x=\frac{5 \pi}{6}\end{align*}x=5π6.

Solution: Substitute in \begin{align*}x=\frac{5 \pi}{6}\end{align*}x=5π6 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*}csc(5π6)21sin(5π6)2112222=0=0=0=0

This is a true statement, so \begin{align*}x=\frac{5 \pi}{6}\end{align*}x=5π6 is a solution to the equation.

Example B

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

Solution: To solve this equation, we need to isolate \begin{align*}\cos x\end{align*}cosx and then use inverse to find the values of \begin{align*}x\end{align*}x 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*}2cosx+12cosxcosx=0=1=12

So, when is the \begin{align*}\cos x=- \frac{1}{2}\end{align*}cosx=12? Between \begin{align*}0 \le x< 2 \pi, x=\frac{2 \pi}{3}\end{align*}0x<2π,x=2π3 and \begin{align*}\frac{4 \pi}{3}\end{align*}4π3. 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*}x=2π3±2πn and \begin{align*}x=\frac{4 \pi}{3} \pm 2 \pi n\end{align*}x=4π3±2πn, where \begin{align*}n\end{align*}n is any integer. You can check your answer graphically by graphing \begin{align*}y=\cos x\end{align*}y=cosx and \begin{align*}y=- \frac{1}{2}\end{align*}y=12 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*}5tan(x+2)1=0, where \begin{align*}0 \le x < 2 \pi\end{align*}0x<2π.

Solution: In this example, we have an interval where we want to find \begin{align*}x\end{align*}x. Therefore, at the end of the problem, we will need to add or subtract \begin{align*}\pi\end{align*}π, 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*}5tan(x+2)15tan(x+2)tan(x+2)=0=1=15

Using the \begin{align*}\tan^{-1}\end{align*}tan1 button on your calculator, we get that \begin{align*}\tan^{-1} \left(\frac{1}{5}\right)=0.1974\end{align*}tan1(15)=0.1974. 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*}

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Date Created:
Apr 23, 2013
Last Modified:
Jul 06, 2016
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