# Question #9440f

Oct 24, 2016

$x \in \left\{\frac{2 \pi}{3} + 4 n \pi , \frac{4 \pi}{3} + 4 n \pi\right\}$

#### Explanation:

We cannot multiply directly to remove the $2$ from the denominator of the argument of sine. Instead, we will let $\theta = \frac{x}{2}$, then solve for $\theta$, and substitute back in after we have gotten rid of the sine function.

Let $\theta = \frac{x}{2}$

$\implies 2 \sqrt{3} \sin \left(\theta\right) = 3$

$\implies \sin \left(\theta\right) = \frac{3}{2 \sqrt{3}} = \frac{\sqrt{3}}{2}$

Checking our unit circle, we find that $\sin \left(\theta\right) = \frac{\sqrt{3}}{2}$ when $\theta = \frac{\pi}{3}$ or $\theta = \frac{2 \pi}{3}$. As sine is periodic with a period of $2 \pi$, we can add any integer multiple $n$ of $2 \pi$ to one of these without changing the value of $\sin e$. So we have:

$\theta = \frac{\pi}{3} + 2 n \pi$

or

$\theta = 2 \frac{\pi}{3} + 2 n \pi$

Now that the sine is removed, let's substitute $\frac{x}{2}$ back in.

$\frac{x}{2} = \frac{\pi}{3} + 2 n \pi$

or

$\frac{x}{2} = \frac{2 \pi}{3} + 2 n \pi$

Multiplying by $2$ we get our desired result:

$x = \frac{2 \pi}{3} + 4 n \pi$

or

$x = \frac{4 \pi}{3} + 4 n \pi$

$\therefore x \in \left\{\frac{2 \pi}{3} + 4 n \pi , \frac{4 \pi}{3} + 4 n \pi\right\}$