Answer
The solution set is $$\{\frac{2\pi}{3}+2n\pi,\frac{4\pi}{3}+2n\pi,n\in Z\}$$
Work Step by Step
$$2\sqrt3\sin\frac{x}{2}=3$$
1) Solve the equation over the interval $[0,2\pi)$
The interval for $x$ is $[0,2\pi)$
As a result, the interval for $\frac{x}{2}$ is $[0,\pi)$
$$2\sqrt3\sin\frac{x}{2}=3$$
$$\sin\frac{x}{2}=\frac{3}{2\sqrt3}=\frac{\sqrt3}{2}$$
Over the interval $[0,\pi)$, there are 2 values of $\frac{x}{2}$ where $\sin\frac{x}{2}=\frac{\sqrt3}{2}$, which are $\{\frac{\pi}{3},\frac{2\pi}{3}\}$
Therefore, $$\frac{x}{2}=\{\frac{\pi}{3},\frac{2\pi}{3}\}$$
We would stop here and not solve for $x$.
2) Solve the equation for all solutions
Sine function has period $2\pi$, so we would add $2\pi$ to all solutions found in part 1) for $\frac{x}{2}$.
$$\frac{x}{2}=\{\frac{\pi}{3}+2n\pi,\frac{2\pi}{3}+2n\pi, n\in Z\}$$
Thus, $$x=\{\frac{2\pi}{3}+2n\pi,\frac{4\pi}{3}+2n\pi,n\in Z\}$$
