Answer
$\frac{\pi}{2},\frac{7\pi}{6},\frac{11\pi}{6}$
Work Step by Step
1. $2sin^2\theta-sin\theta-1=0 \Longrightarrow (sin\theta-1)(2sin\theta+1)=0 \Longrightarrow sin\theta=-\frac{1}{2},1$,
2. $sin\theta=1 \Longrightarrow \theta=\frac{\pi}{2}$ in $[0,2\pi)$,
3. $sin\theta=-\frac{1}{2} \Longrightarrow \theta=\frac{7\pi}{6},\frac{11\pi}{6}$ in $[0,2\pi)$.
