Answer
$\pi$
Work Step by Step
1. $sin^2\theta=6(cos(-\theta)+1) \Longrightarrow 1-cos^2\theta=6cos\theta+6 \Longrightarrow cos^2\theta+6cos\theta+5=0 \Longrightarrow (cos\theta+1)(cos\theta+5)=0 \Longrightarrow cos\theta=-5,-1$,
2. $cos\theta=-5 \Longrightarrow$ no solution in $[0,2\pi)$,
3. $cos\theta=-1 \Longrightarrow \theta=\pi$ in $[0,2\pi)$.
