Answer
$\{\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6},\frac{3\pi}{2} \}$
Work Step by Step
Step 1. Use the identity $sin(2x)=2sin(x)cos(x)$, we have $cos(x)(2sin(x)-1)=0$, which gives $cos(x)=0$ and $sin(x)=\frac{1}{2}$
Step 2. For $cos(x)=0$, we can find all x-values in $[0,2\pi)$ as $x=\frac{\pi}{2},\frac{3\pi}{2}$
Step 3. For $sin(x)=\frac{1}{2}$ , we can find all x-values in $[0,2\pi)$ as $x=\frac{\pi}{6},\frac{5\pi}{6}$
Step 4. The solutions for the original equation in $[0,2\pi)$ are $\{\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6},\frac{3\pi}{2} \}$
