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