Answer
$\frac{3\pi}{2},\frac{\pi}{6},\frac{5\pi}{6}$
Work Step by Step
1. $1+sin\theta=2cos^2\theta \Longrightarrow 1+sin\theta=2-2sin^2\theta \Longrightarrow 2sin^2\theta+sin\theta-1=0 \Longrightarrow (2sin\theta-1)(sin\theta+1)=0 \Longrightarrow sin\theta=-1\ or\ sin\theta=\frac{1}{2}$,
2. $sin\theta=-1 \Longrightarrow \theta=\frac{3\pi}{2}$ in $[0,2\pi)$,
3. $sin\theta=\frac{1}{2} \Longrightarrow \theta=\frac{\pi}{6},\frac{5\pi}{6}$ in $[0,2\pi)$.
