Answer
$\frac{3\pi}{2}$
Work Step by Step
1. $4(1+sin\theta)=cos^2\theta \Longrightarrow 4+4sin\theta=1-sin^2\theta \Longrightarrow sin^2\theta+4sin\theta+3=0 \Longrightarrow (sin\theta+3)(sin\theta+1)=0 \Longrightarrow sin\theta=-3,-1$,
2. $sin\theta=-3 \Longrightarrow $ no solution in $[0,2\pi)$,
3. $sin\theta=-1 \Longrightarrow \theta=\frac{3\pi}{2}$ in $[0,2\pi)$.
