Answer
$0,\frac{\pi}{3},\frac{5\pi}{3}$
Work Step by Step
1. $2sin^2\theta-3(1-cos(-\theta)) \Longrightarrow 2-2cos^2\theta=3-3cos\theta \Longrightarrow 2cos^2\theta-3cos\theta+1=0 \Longrightarrow (cos\theta-1)(2cos\theta-1)=0 \Longrightarrow cos\theta=\frac{1}{2},1$,
2. $cos\theta=1 \Longrightarrow \theta=0$ in $[0,2\pi)$,
3. $cos\theta=\frac{1}{2} \Longrightarrow \theta=\frac{\pi}{3},\frac{5\pi}{3}$ in $[0,2\pi)$.
