<s>[INST] 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 $12=\sin(\theta)+2\sin^2(\theta)$
 [/INST] Note that
\begin{eqnarray*}
12\sin \theta  &=&\cos (2\theta ) \\
&=&1-2\sin ^{2}\theta 
\end{eqnarray*}
then
\begin{equation*}
2\sin ^{2}\theta +12\sin \theta -1=0.
\end{equation*}
Let $X=\sin \theta ,$ then
\begin{equation*}
2X^{2}+12X-1=0
\end{equation*}
This quadratic equation have two roots: $X_{1}=\frac{1}{2}\sqrt{38}-3$ and $%
X_{2}=-\frac{1}{2}\sqrt{38}-3.$ Note that $X_{2}<-3$ then $X_{2}\notin \left[
-1,1\right] ,$ then it cannot provide a $\theta -solution$ to the original
equation. Next, $\frac{1}{2}\sqrt{38}-3\simeq 0.082\,2\in \left[ -1,1\right] 
$ then one has
\begin{equation*}
\sin \theta =\frac{1}{2}\sqrt{38}-3
\end{equation*}
and then
\begin{equation*}
\theta =\arcsin \left( \frac{1}{2}\sqrt{38}-3\right) \simeq 0.0823\ rad
\end{equation*}
Remark. It is known that when $\theta $ (in rad) is so small then $\sin \theta
\simeq \theta .$ Since our $\sin \theta =X_{1}\simeq 0.082\,2$ is so small
then it is not surprising to find $\theta \simeq \sin \theta .$
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