Solving $2\cos\left(2\theta\right) = \sqrt{3}$ I have a question on this test review problem (that will help us on a test), and I have no clue what it's asking. We're learning trigonometry, (Analytic Trigonometry), like about the unit circle, inverse trig functions, etc... And I encounter this problem on test review:

$$ 2\cos\left(2\theta\right) =  \sqrt{3}$$

I know the answers are: $\left\{\frac{\pi}{12}, \frac{11\pi}{12}, \frac{13\pi}{12}, \frac{23\pi}{12} \right\} $ 
But I want to know how to find it. So I get ready for test. Thank you.
EDIT
After some thinking i got some of it but not All!! So what i did was:
$ 2\Theta = \cos^{-1} (\frac{\sqrt3}{2} )$
$2\Theta = \frac {\pi}{6}$
$\Theta = \frac{\pi}{12} + 2\pi n$
So now i just add 2pi but i have to remember $0 \le  \Theta \lt 2\pi$
I get: 
$\left\{\frac{\pi}{12}, \frac{13\pi}{12} \right\} $ But that obviously isnt the anwser.. Please help!
 A: This writes as $\;\cos2\theta=\dfrac{\sqrt 3}2=\cos\dfrac\pi6$. Now you have to know that


*

*$\cos x=\cos\alpha\iff\theta\equiv\pm\alpha\pmod2\pi$, and similarly:

*$\sin  x=\sin \alpha\iff\begin{cases}\theta\equiv\alpha\pmod{2\pi},\\\theta\equiv\pi-\alpha\pmod{2\pi},
\end{cases}$

*$\tan x=\tan\alpha\iff x\equiv\alpha\pmod\pi.$


So here, you have
$$2\theta\equiv \pm\frac\pi6\pmod{2\pi}\iff \theta\equiv\frac\pi{12}\pmod{\pi}$$
If you want the solutions in $[0,2\pi]$, you get
$$\theta =\frac\pi{12},-\frac\pi{12}+\pi=\frac{11\pi}{12},\frac\pi{12}+\pi=\frac{13\pi}{12},-\frac\pi{12}+2\pi=\frac{23\pi}{12}.$$
A: I would reform the equation $2\cos 2\theta = \sqrt{3}$ into $$\cos 2\theta \ =\frac{\sqrt 3}{2}$$
For now, I wouldn't worry about the $2\theta$ because if $\theta$ is an angle, then so is $2\theta$, so my intermediary strategy is to find $2\theta$. Later I can always divide that number by 2.
So lets fund $2\theta$.
I'm looking for a triangle that has an adjacent side of $\sqrt 3$ and a hypotenuse of $2$. If I take an  equilateral triangle, and split it in half, I will have such a triangle.

So I know that that the angle that I'm looking for is $30^\circ$, or $\frac{\pi}{6}$ rads. In other words, $2\theta = \frac{\pi}{6}$, solving for $\theta$ we have $\theta = \frac{\pi}{12}$.
Now that you have your reference angle all figured out, try to figure out what quadrants will give you a positive cosine. I think that you will find that the angles are $\theta = (2\pi)n\pm\frac{\pi}{12}$
A: Perhaps the included picture will help you understand the solutions.. The figure shows $\cos \theta$ in the range $-2 \pi \le \theta \le 2 \pi$. The blue line is the value $\cos \frac{\pi}{12}$ and you can see that it cuts the curve at four values in the request range. It's difficult to make out the required solutions (but you know them anyway), but it may help you understand whey there are four answers.
A: we have $\cos \frac{\pi}{6} = \frac{\sqrt{3}}2$. Hence your equation reads
$$  \cos 2\theta = \cos \frac{\pi}{6}$$
then $$2\theta = \frac{\pi}{6} + 2k\pi \ \ \mbox{ or }  2\theta = -\frac{\pi}{6} + 2k\pi \ \  \ \ \ \ \ (k\in\mathbb{Z})$$
that is 
$$\theta = \frac{\pi}{12} + k\pi \ \ \mbox{ or }  \theta = -\frac{\pi}{12} + k\pi \ \  \ \ \ \ \ (k\in\mathbb{Z})$$
It is clear that there are an infinite number of solution not only 4. (But if you precise an appropriate interval for solutions, you can make it to have those solutions in your question  )
