Answer
$\frac{\pi}{2},\frac{3\pi}{2},\frac{7\pi}{6},\frac{11\pi}{6}$
Work Step by Step
1. $tan(2\theta)+2cos\theta=0 \Longrightarrow \frac{sin(2\theta)}{cos(2\theta)}+2cos\theta=0 \Longrightarrow \frac{2sin\theta cos\theta}{1-2sin^2\theta}+2cos\theta=0\Longrightarrow cos\theta=0\ or\ \frac{sin\theta}{1-2sin^2\theta}+1=0$
2. For $cos\theta=0$, we have $\theta=k\pi+\frac{\pi}{2}$,
3. $\frac{sin\theta}{1-2sin^2\theta}+1=0\Longrightarrow 2sin^2\theta-sin\theta-1\Longrightarrow (2sin\theta+1)(sin\theta-1)=0\Longrightarrow sin\theta=1, -\frac{1}{2}$,
4. For $sin\theta=-1$, we have $\theta=2k\pi+\frac{3\pi}{2}$
5. For $sin\theta=-\frac{1}{2}$, we have $\theta=2k\pi+\frac{7\pi}{6}$ or $2k\pi+\frac{11\pi}{6}$,
6. Within $[0,2\pi)$, we have $\theta=\frac{\pi}{2},\frac{3\pi}{2},\frac{7\pi}{6},\frac{11\pi}{6}$
