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# 14.3: Solving Trigonometric Equations

Difficulty Level: At Grade Created by: CK-12

Objective

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

Review Queue

Solve the following equations.

1. $x^2-7x-18=0$

2. $9x^4-16=0$

3. $\sin x= \frac{\sqrt{2}}{2},0< x < \frac{\pi}{2}$

## Using Algebra

Objective

Here you'll solve trig equations using algebra.

Guidance

In the previous concept, we verified trigonometric identities, which are true for every real value of $x$. In this concept, we will solve trigonometric equations. An equation is only true for some values of $x$.

Example A

Verify that $\csc x-2=0$ when $x=\frac{5 \pi}{6}$.

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

$\csc \left(\frac{5 \pi}{6}\right)-2&=0 \\\frac{1}{\sin \left(\frac{5 \pi}{6}\right)}-2&=0 \\\frac{1}{\frac{1}{2}}-2&=0 \\2-2&=0$

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

Example B

Solve $2 \cos x+1=0$.

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

$2 \cos x+1&=0 \\2 \cos x&=-1 \\\cos x&=- \frac{1}{2}$

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

Example C

Solve $5 \tan(x+2)-1=0$, where $0 \le x < 2 \pi$.

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

$5 \tan(x+2)-1&=0 \\5 \tan(x+2)&=1 \\\tan(x+2)&=\frac{1}{5}$

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

$x+2&=0.1974 \\x&=-1.8026$

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

$x&=-1.8026+ \pi=1.3390 \\&=1.3390+ \pi=4.4806$

The two solutions are $x = 1.3390$ and 4.4806.

Guided Practice

1. Determine if $x=\frac{\pi}{3}$ is a solution for $2 \sin x=\sqrt{3}$.

Solve the following trig equations in the interval $0 \le x< 2 \pi$.

2. $3 \cos^2x-5=0$

3. $3 \sec(x-1)+2=0$

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

2. Isolate the $\cos^2x$ and then take the square root of both sides. Don’t forget about the $\pm$!

$9 \cos^2x-5&=0 \\9 \cos^2x&=5 \\\cos^2x&=\frac{5}{9} \\\cos x&=\pm \frac{\sqrt{5}}{3}$

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

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

3. Here, we will find the solution within the given range, $0 \le x< 2 \pi$.

$3 \sec(x-1)+2&=0 \\3 \sec(x-1)&=-2 \\\sec(x-1)&=- \frac{2}{3} \\\cos(x-1)&=- \frac{3}{2}$

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

Problem Set

Determine if the following values for $x$. are solutions to the equation $5+6 \csc x=17$.

1. $x=- \frac{7 \pi}{6}$
2. $x=\frac{11 \pi}{6}$
3. $x=\frac{5 \pi}{6}$

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

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

Sole the following trigonometric equations within the interval $0 \le x < 2 \pi$. If no solutions exist, write no solution.

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

## By Using Quadratic Techniques

Objective

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

Guidance

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 $\sin^2x-3 \sin x+2=0$.

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

$\sin^2x-3 \sin x+2&=0 \\(\sin x-2)(\sin x-1)&=0 \\\sin x=2 \ and \ \sin x&=1$

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

Example B

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

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

$1- \sin x&=\sqrt{3} \cdot \sqrt{1- \sin^2x} \\(1- \sin x)^2&=\sqrt{3-3 \sin^2x}^2 \\1-2 \sin x+\sin^2x&=3-3 \sin^2x \\4 \sin^2x-2 \sin x-2&=0 \\2 \sin^2x-\sin x-1&=0 \\(2 \sin x+1)(\sin x-1)&=0$

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

Example C

Solve $\tan^2x-5 \tan x-9=0$ in the interval $0 \le x < \pi$.

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

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

$x \approx \tan^{-1} 6.41 \approx 1.416 \ \text{rad}$ and $x \approx \tan^{-1}-1.41 \approx -0.954 \ \text{rad}$

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, $\pi - 0.954=2.186 \ \text{rad}$.

Guided Practice

Solve the following trig equations using any method in the interval $0 \le x < 2 \pi$.

1. $\sin^2x \cos x=\cos x$

2. $\sin^2x=2 \sin(-x)+1$

3. $4 \cos^2x-2 \cos x-1=0$

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

$\sin^2x \cos x- \cos x&=0 \\\cos x(\sin^2x-1)&=0 \\\cos x(\sin x -1)(\sin x+1)&=0$

$\cos x&=0 \qquad \qquad \ \ \sin x=1 \qquad \ \ \sin x=-1 \\x&=\frac{\pi}{2} \ and \ \frac{3 \pi}{2} \qquad \ x=\frac{\pi}{2} \qquad \qquad x=\frac{3 \pi}{2}$

2. Recall that $\sin(-x)=- \sin x$ from the Negative Angle Identities.

$\sin^2x&=2 \sin(-x)+1 \\\sin^2x&=-2 \sin x+1 \\\sin^2x+2 \sin x+1&=0 \\(\sin x+1)^2&=0 \\\sin x&=-1 \\x&=\frac{3 \pi}{2}$

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

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

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

The other solutions in the range are $x \approx 2 \pi - 0.6283 \approx 5.6549$ and $x \approx \pi + 1.2570 \approx 4.3982$.

Problem Set

Solve the following trig equations using any method. Find all solutions in the interval $0 \le x < 2 \pi$. Round any decimal answers to 4 decimal places.

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

Using your graphing calculator, graph the following equations and determine the points of intersection in the interval $0 \le x < 2 \pi$.

1. .
$y&=\sin^2x \\y&=2 \sin x-1$
1. .
$y&=4 \cos x-3 \\y&=-2 \tan x$

## Date Created:

Apr 23, 2013

Dec 16, 2014
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