Answer
$0,\pi, \frac{\pi}{3},\frac{5\pi}{3}$
Work Step by Step
1. Let $f(x)=0$ or $sin(2x)-sin(x)=0\Longrightarrow 2sin(x)cos(x)-sin(x)=0\Longrightarrow sin(x)=0$ or $cos(x)=\frac{1}{2}$
2. For $sin(x)=0$, we have $x=k\pi$.
3. For $cos(x)=\frac{1}{2}$, we have $x=2k\pi+\frac{\pi}{3}$ or $2k\pi+\frac{5\pi}{3}$,
4. Within $[0,2\pi)$, we have $\theta=0,\pi, \frac{\pi}{3},\frac{5\pi}{3}$
