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