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