Prove trigonometric identity, hence or otherwise find the general solution The following question requires one to prove the below trigonometric identity $$\cos 3x = 4\cos ^3 x - 3\cos x$$
Hence, or otherwise, find the general solution of the following equation $$(4\cos ^2 x - 3)(4\cos ^2 3x - 3)(4\cos ^2 9x -3) = 1$$

After proving the latter over a couple back and forth steps (I'll be honest, I took longer than I should've), I've proceeded to the second part of the question; key term in the question I thought is hence. Thus I had to somehow use the previous proof and simplify the above equation.
I attempted to make the relation between $\cos3x$ and $4\cos ^2 3x -3$ however can't seem to break anything valuable.

I would greatly appreciate any hints that could help me move forward with this question, is there an expansion that I am to be made aware of; does $2n\pi \pm \alpha$ take place in the solution?
 A: We start with
\begin{align*}
4\cos ^2 x - 3)(4\cos ^2 3x - 3)(4\cos ^2 9x -3) & = 1\\
\cos x (4\cos ^2 x - 3)(4\cos ^2 3x - 3)(4\cos ^2 9x -3) & = \cos x\\
(4\cos ^3 x - 3\cos x)(4\cos ^2 3x - 3)(4\cos ^2 9x -3) & = \cos x\\
(\cos 3x)(4\cos ^2 3x - 3)(4\cos ^2 9x -3) & = \cos x\\
(4\cos ^3 3x - 3\cos 3x)(4\cos ^2 9x -3) & = \cos x\\
\cos 9x (4\cos ^2 9x -3) & = \cos x\\
\cos 27 x & = \cos x.
\end{align*}
Now solve.
A: Verify the following identity:$\cos 3x = 4\cos ^3 x - 3\cos x$
$$\cos (3x) = 4\cos ^3 (x) - 3\cos (x)\Longleftrightarrow$$
$$\cos (3x) = 4\left(\frac{1}{4}(3\cos(x)+\cos(3x))\right) - 3\cos (x)\Longleftrightarrow$$
$$\cos(3x)=4\left(\frac{3\cos(x)+\cos(3x)}{4}\right)-3\cos(x)\Longleftrightarrow$$
$$\cos(3x)=\frac{4(3\cos(x)+\cos(3x))}{4}-3\cos(x)\Longleftrightarrow$$
$$\cos(3x)=\frac{4}{4}(3\cos(x)+\cos(3x))-3\cos(x)\Longleftrightarrow$$
$$\cos(3x)=3\cos(x)+\cos(3x)-3\cos(x)\Longleftrightarrow$$
$$\cos(3x)=3\cos(x)-3\cos(x)+\cos(3x)\Longleftrightarrow$$
$$\cos(3x)=0+\cos(3x)\Longleftrightarrow$$
$$\cos(3x)=\cos(3x)$$
Q.E.D.
