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 $12=\sin(\theta)+2\sin^2(\theta)$
 A: 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?
A: 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?).
A: you are almost there. $$2\sin^2 \theta + \sin \theta - 12=0 $$ so $$\sin \theta = \frac{-1 \pm \sqrt{49}}{4} = -2, \frac32$$ neither of them leads to a solution for $\theta.$ 
we should have seen this from the equation itself. reason is $$1 \ge |\cos 2\theta| = |12 - \sin \theta | \ge 11$$ a contradiction.
A: 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 $\theta \simeq \sin \theta .$
A: $12-\sin(\theta)$ oscillates between 11 and 13. 
$\cos(2\theta)$oscillates between 1 and -1.
$ 11> 1. $
The two graphs have no cutting point, so no real roots solution at all, but complex roots are: 
$ \sin^{-1}(-2), \sin^{-1}{} \frac32. $
A: From

$$12-\sin(\theta)=\cos(2\theta)$$

you get
$$12-\sin(\theta)=\cos^2(\theta)-\sin^2(\theta),$$
this is
$$12-\sin(\theta)=1-2\sin^2(\theta).$$
Hence
$$2\sin^2(\theta)-\sin(\theta)+11=0$$
and
$$\sin(\theta)=\frac{1\pm\sqrt{1-4\cdot2\cdot11}}{4}=\frac{1\pm\sqrt{-87}}{4}.$$
So, no real solution for $\sin(\theta)$ means no real solution for $\theta$ and, in particular, no solution $\theta\in[0,2\pi)$.
Indeed, $12-\sin(\theta)\in[11,13]$ while $\cos(2\theta)\in[-1,1]$, for all $\theta\in\Bbb R$.
A: $$12-\sin(x)=\cos(2x)\Longleftrightarrow$$
$$12-\cos(2x)-\sin(x)=0\Longleftrightarrow $$
$$11-\sin(x)+2\sin^2(x)=0\Longleftrightarrow $$
$$\frac{11}{2}-\frac{\sin(x)}{2}+\sin^2(x)=0\Longleftrightarrow $$
$$\sin^2(x)-\frac{\sin(x)}{2}=-\frac{11}{2}\Longleftrightarrow $$
$$\frac{1}{16}-\frac{\sin(x)}{2}+\sin^2(x)=-\frac{87}{16}\Longleftrightarrow $$
$$\left(\sin(x)-\frac{1}{4}\right)^2=-\frac{87}{16}\Longleftrightarrow $$
$$\sin(x)-\frac{1}{4}=\pm \frac{i\sqrt{87}}{4}\Longleftrightarrow $$
$$\sin(x)=\pm \frac{i\sqrt{87}}{4}+\frac{1}{4}\Longleftrightarrow $$
So now the solutions:
$$x_1=2\left(\pi n+\tan^{-1}\left(\frac{1}{22}-\frac{i\sqrt{87}}{22}\pm \frac{1}{2}\sqrt{-\frac{570}{121}-\frac{2i\sqrt{87}}{121}}
\right)\right)$$
$$x_2=2\left(\pi n+\tan^{-1}\left(\frac{1}{22}+\frac{i\sqrt{87}}{22}\pm \frac{1}{2}\sqrt{-\frac{570}{121}-\frac{2i\sqrt{87}}{121}}
\right)\right)$$
With $n$ is the element of Z -> the set of integers
