How to factorize $\frac{\cos(3x)-\cos(x)}{\tan(2x)-\tan(x)}$? How to factorize $\dfrac{\cos(3x)-\cos(x)}{\tan(2x)-\tan(x)}$?
Which trigonometric identities to use?
I'm stuck when it comes to $\tan(2x)+\tan(x)$. I don't know which identity to use to turn it into the product.
I was thinking of just transforming them to sines and cosines, but also doesn't get me anywhere.
Thanks for help in advance.

First note the following identities:
\begin{align}
\cos (3x) &= 4 \cos ^3 (x) - 3 \cos (x) \\
\sin(2x) &= 2 \sin(x) \cos(x) \\
\cos(2x) &= 2 \cos^2(x) - 1
\end{align}
Substituting into the expression yields
\begin{multline}
\frac{\cos(3x) - \cos(x)}{\tan(2x) - \tan(x)} = \frac{4 \cos(x)(\cos^2(x) - 1)}{\frac{2 \sin(x) \cos(x)}{2 \cos^2(x) - 1} - \frac{\sin(x)}{\cos(x)}} = \frac{4 \cos^2(x)(\cos^2(x) - 1)(2\cos^2(x) - 1)}{2 \sin(x) \cos^2(x) - \sin(x)(2\cos^2(x) - 1)} \\= 4 \frac{\cos^2(x)}{\sin(x)}(\cos^2(x) - 1)(2\cos^2(x) - 1).
\end{multline}
Now, substituting back in $2 \cos^2(x) - 1=\cos(2x)$ and $\cos^2(x) - 1 = -\sin^2(x)$ yields
\begin{multline}4 \frac{\cos^2(x)}{\sin(x)}(\cos^2(x) - 1)(2\cos^2(x) - 1) = -4 \cos^2(x) \sin(x) \cos(2x)\\  = -2\cos(x)\sin(2x)\cos(2x) = -\cos(x)\sin(4x)\end{multline}
where in the last two equalities I have used $2 \sin(x) \cos(x) = \sin(2x)$.
