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