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