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 $\cos(2a) = 4\cos^2(3a)-1$
 A: $$2\cos3\alpha+2\cos\alpha=3\cos\alpha$$ or
$$4\cos2\alpha\cos\alpha=3\cos\alpha.$$
Can you end it now?
I got $\cos2\alpha=-1$ or $\cos2\alpha=\frac{3}{4}.$
I used the following identity.
$$\cos\alpha+\cos\beta=2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2},$$ which we can get by the following way.
$$\cos\alpha+\cos\beta=\cos\left(\frac{\alpha+\beta}{2}+\frac{\alpha-\beta}{2}\right)+\cos\left(\frac{\alpha+\beta}{2}-\frac{\alpha-\beta}{2}\right)=$$
$$=\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}-\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}+$$
$$+\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}+\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}=$$
$$=2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}.$$
We can use also the following way.
$$2\cos(2\alpha+\alpha)=\cos\alpha$$ or
$$2(\cos2\alpha\cos\alpha-\sin2\alpha\sin\alpha)=\cos\alpha$$ or
$$2(\cos2\alpha\cos\alpha-2\sin^2\alpha\cos\alpha)=\cos\alpha.$$ 
Now, if $\cos\alpha=0$ we obtain $$\cos2\alpha=-1.$$
Otherwise, we obtain 
$$2(\cos2\alpha-2\sin^2\alpha)=1$$ or
$$2(\cos2\alpha-1+\cos2\alpha)=1,$$ which gives
$$\cos2\alpha=\frac{3}{4}.$$
A: $$2\cos3a=\cos a$$
$$4\cos^23a=\cos^2a$$
$$4(1+\cos6a)=1+\cos2a$$
$$4+4(4\cos^32a-3\cos2a)=1+\cos2a$$
with $x=\cos2a$ then
$$16x^3-13x+3=0$$
$$13(x^3-x)+3(x^3+1)=0$$
$$(x+1)(13x^2-13x+3x^2-3x+3)=0$$
$$(x+1)(16x^2-16x+3)=0$$
tis gives the answers $\cos2a=-1$ or $\cos2a=\dfrac14$ or $\cos2a=\dfrac34$.
A: Hint:
$$a\cos3A=b\cos A$$
$$\iff a(4\cos^3A-3\cos A)=b\cos A$$
$$\iff\cos A[4a\cos^2A-3a-b]=0$$
Use $\cos2A=2\cos^2A-1$
A: 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

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

