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 $12=\sin(\theta)+2\sin^2(\theta)$

Note that $-1\le \sin \theta \le 1\implies 11\le 12-\sin \theta \le 13$
Also $-1\le\cos \theta \le 1\implies -1\le \cos 2\theta \le 1$
Can you take it from there?
