Answer
$\{\frac{\pi}{6},\frac{\pi}{3},\frac{7\pi}{6},\frac{4\pi}{3} \}$
Work Step by Step
Step 1. Using the identity $sin(2x)=2sin(x)cos(x)$, we have $sin(2x)=\frac{\sqrt3}{2}$, which gives $2x=2k\pi+\frac{\pi}{3}$ and $2x=2k\pi+\frac{2\pi}{3}$ or $x=k\pi+\frac{\pi}{6}$ and $x=k\pi+\frac{\pi}{3}$ where $k$ is an integer.
Step 2. For $x=k\pi+\frac{\pi}{6}$, we can find all x-values in $[0,2\pi)$ as $x=\frac{\pi}{6},\frac{7\pi}{6}$
Step 3. For $x=k\pi+\frac{\pi}{3}$, we can find all x-values in $[0,2\pi)$ as $x=\frac{\pi}{3},\frac{4\pi}{3}$
Step 4. The solutions for the original equation in $[0,2\pi)$ are $\{\frac{\pi}{6},\frac{\pi}{3},\frac{7\pi}{6},\frac{4\pi}{3} \}$
