Answer
$0,\frac{\pi}{6},\pi, \frac{11\pi}{6}$.
Work Step by Step
Step 1. Rewrite the equation as $2sin(x)cos(x)=\sqrt 3sin(x)$ or $sin(x)(2cos(x)-\sqrt 3)=0$
Step 2. For $sin(x)=0$, we have solutions in $[0,2\pi)$ as $x=0,\pi$
Step 3. For $cos(x)=\frac{\sqrt 3}{2}$, we have solutions in $[0,2\pi)$ as $x=\frac{\pi}{6}, \frac{11\pi}{6}$
Step 4. Thus, we have all the solutions in $[0,2\pi)$ as $x=0,\frac{\pi}{6},\pi, \frac{11\pi}{6}$.
