How to solve $12-\sin(\theta)=\cos(2\theta)$? $$12-\sin(\theta)=\cos(2\theta)$$
What's the correct answer on the $[0,2\pi]$? 
I started with  $12-\sin(\theta)=1-2\sin^2(\theta)$ and then i cant get anything sensible as i end up with $12=\sin(\theta)+2\sin^2(\theta)$

Hint: So far so good: $$2\sin^2\theta + \sin\theta - 12= 0 \implies \sin \theta = \frac{-1\pm\sqrt{1^2-4\cdot 2 \cdot(-12)}}{2\cdot 2}.$$
Find two possible values for $\sin \theta$. Then find the possible values for $\theta$ (will there be any?).
