# 3.2.6: Solving Trigonometric Equations Using Basic Algebra


Substitute in potential solutions or solve using inverse trig functions.

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

## Solving Trigonometric Equations

We have already 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 $$x$$.

Let's verify that $$\csc x−2=0$$ when $$x=\dfrac{5\pi}{6}$$.

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

\begin{aligned} \csc \left(\dfrac{5\pi}{6}\right)−2&=0\\ \dfrac{1}{\sin \left(\dfrac{5\pi}{6}\right)}−2&=0 \\ \dfrac{1}{\dfrac{1}{2}}−2&=0 \\ 2−2&=0\end{aligned}

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

Now, let's solve $$2\cos x+1=0$$.

To solve this equation, we need to isolate cosx and then use inverse to find the values of x when the equation is valid.

\begin{aligned} 2\cos x+1&=0 \\ 2\cos x&=−1 \\ \cos x&=−\dfrac{1}{2}\end{aligned}

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

Finally, let's solve $$5 \tan(x+2)−1=0$$, where $$0\leq x<2\pi$$.

In this problem, 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.

\begin{aligned} 5 \tan(x+2)−1&=0 \\ 5 \tan(x+2)&=1 \\ \tan(x+2)&=\dfrac{1}{5} \end{aligned}

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

\begin{aligned} x+2&=0.1974 \\ x&=−1.8026 \end{aligned}

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 $$[0,2\pi ]$$.

\begin{aligned} x&=−1.8026+\pi =1.3390 \\ &=1.3390+\pi =4.4806\end{aligned}

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

##### Example $$\PageIndex{1}$$

Earlier, you were asked to find the value of x from the equation $$2\sin x−\sqrt{2}=0$$.

Solution

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.

\begin{aligned} 2 \sin x−\sqrt{2}&=0 \\ 2\sin x&=\sqrt{2} \\ \sin x&=\dfrac{\sqrt{2}}{2}\end{aligned}

So now we need to find the values of $$x$$ for which $$\sin x=\dfrac{\sqrt{2}}{2}$$. 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 $$0\leq x<2\pi$$. Recall that the sine is positive in both the first and second quadrants, so $$\sin x=\dfrac{\sqrt{2}}{2}$$ when $$x$$ also is $$135^{\circ}$$.

##### Example $$\PageIndex{2}$$

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

Solution

$$2\sin \dfrac{\pi}{3}=\sqrt{3} \rightarrow 2 \cdot \dfrac{\sqrt{3}}{2}=\sqrt{3}$$

Yes, $$x=\dfrac{\pi}{3}$$ is a solution.

##### Example $$\PageIndex{3}$$

Solve the following trig equation in the interval $$0\leq x<2\pi$$.

$$3 \cos^2 x−5=0$$

Solution

\begin{aligned} 9\cos^2x−5&=0 \\ 9\cos^2x&=5 \\ \cos^2 x&=\dfrac{5}{9} \\ \cos x=\pm \dfrac{\sqrt{5}}{3}\end{aligned}

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

The $$\cos x=−\dfrac{\sqrt{5}}{3}$$ at $$x=2.412 \text{rad}$$, 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 $$\pi$$. $$x=\pi +0.243=3.383\text{ rad}$$.

##### Example $$\PageIndex{4}$$

Solve the following trig equation in the interval $$0\leq x<2\pi$$.

$$3\sec(x−1)+2=0$$

Solution

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

\begin{aligned} 3\sec(x−1)+2&=0 \\ 3\sec(x−1)&=−2 \\ \sec(x−1)&=−\dfrac{2}{3} \\ \cos(x−1)&=−\dfrac{3}{2}\end{aligned}

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

## Review

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

1. $$x=−\dfrac{7\pi }{6}$$
2. $$x=\dfrac{11\pi }{6}$$
3. $$x=\dfrac{5\pi }{6}$$

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 $$0\leq x<2\pi$$. 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. $$5+4\cot^2 x=17$$
6. $$2\sin^2x−7=−6$$