Answer
$x=4k\pi+\frac{2\pi}{3}$ and $x=4k\pi+\frac{4\pi}{3}$;
or $x=720^\circ k+120^\circ$ and $x=720^\circ k+240^\circ$ where $k$ is an integer.
Work Step by Step
1. Given $2\sqrt 3 sin\frac{x}{2}=3$, we have $sin\frac{x}{2}=\frac{\sqrt 3}{2}$
2. The period is $4\pi$ and the fundamental solutions (reference angles) with $\frac{x}{2}\in[0,2\pi)$ are $\frac{x}{2}=\frac{\pi}{3},\frac{2\pi}{3}\longrightarrow x=\frac{2\pi}{3},\pi+\frac{4\pi}{3}$, thus the general solutions are $x=4k\pi+\frac{2\pi}{3}$ and $x=4k\pi+\frac{4\pi}{3}$ where $k$ is an integer.
3. We can also write the solutions as $x=720^\circ k+120^\circ$ and $x=720^\circ k+240^\circ$ where $k$ is an integer.
