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# Solving Trigonometric Equations Using Basic Algebra

## Substitute in potential solutions or solve using inverse trig functions.

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Practice Solving Trigonometric Equations Using Basic Algebra
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Solving Trigonometric Equations Using Algebra

As Agent Trigonometry, you are given this clue: $2 \sin x-\sqrt{2}=0$ . If $0 \le x < 2 \pi$ , what is/are the value(s) of $x$ .

### 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.

Concept Problem Revisit To solve this equation, we need to isolate $\sin x$ and then use inverse to find the values of $x$ when the equation is valid.

$2 \sin x -\sqrt{2}=0 \\2 \sin x&=\sqrt{2} \\\sin x&=\frac{\sqrt{2}}{2}$

So now we need to find the values of $x$ for which $\sin x =\frac{\sqrt{2}}{2}$ . We know from the special triangles that this value of sine holds true for a $45^\circ$ angle, but is that the only value of $x$ for which it is true?

We are told that $0 \le x < 2 \pi$ . Recall that the sine is positive in both the first and second quadrants, so $\sin x =\frac{\sqrt{2}}{2}$ when $x$ also is $135^\circ$ .

### 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 $2^{nd}$ quardrant. 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.

### Practice

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. $2 \sin^2x-7=-6$