Trigonometric Identities: Given that $2\cos(3a)=\cos(a)$ find $\cos(2a)$ 
Given that $2\cos(3a)=\cos(a)$ find $\cos(2a)$.

$2\cos(3a)=\cos(a)$
I converted $\cos(2a)$ into $\cos^2(a)-\sin^2(a)$
Then I tried plugging in. I know this is not right, but I have no clue how to solve this. Hints please?
edit: 
Because I got that $\cos(2a) = 4\cos^2(3a)-1$

I like to split $\color{blue}{\cos(3 a) = \cos(a+2a) = \cos(a) \cos(2 a)  -\sin(a) \sin(2 a)}$. This makes the expression
$$ 2 \left( \cos(a) \cos(2 a)  -\sin(a) \sin(2 a) \right) = \cos(a) $$
or (re-arrange and divide by $2\cos(a)$) 
$$ \frac{2 \cos(a) \cos(2 a)}{2 \cos(a)} = \frac{\cos(a)}{2 \cos(a)} + \frac{2\sin(a) \sin(2 a)}{2 \cos(a) }$$
of course $\color{blue}{ \sin(2 a) = 2\sin(a)\cos(a)}$
$$ \cos(2 a) = \tfrac{1}{2} + \frac{ 2\sin(a) \left( 2 \sin(a) \cos(a) \right)}{2 \cos(a)} = \tfrac{1}{2} + 2 \sin^2(a) $$
Now use $\color{blue}{\sin^2(a) = \tfrac{1}{2} - \tfrac{1}{2} \cos(2 a)}$ to make the above
$$ \cos(2 a) = \tfrac{1}{2} + 2 \left(\tfrac{1}{2} - \tfrac{1}{2} \cos(2 a)\right) = \tfrac{3}{2} - \cos(2 a)  $$
The above is solved for

$$ \cos(2 a) = \tfrac{3}{4} $$

