Answer
$\frac{\pi}{2},\frac{3\pi}{2},\frac{2\pi}{3},\frac{4\pi}{3}$
Work Step by Step
1. $sin^2\theta-cos^2\theta=1+cos\theta \Longrightarrow 1-2cos^2\theta=1+cos\theta \Longrightarrow 2cos^2\theta+cos\theta=0 \Longrightarrow cos\theta=-\frac{1}{2},0$,
2. $cos\theta=0 \Longrightarrow \theta=\frac{\pi}{2},\frac{3\pi}{2}$ in $[0,2\pi)$,
3. $cos\theta=-\frac{1}{2} \Longrightarrow \theta=\frac{2\pi}{3},\frac{4\pi}{3}$ in $[0,2\pi)$.
