Show that $ \frac{\cos5x + \cos4x} {1-2\cos3x} = -\cos2x -\cos x $ The Question reads - 
$$ \frac{\cos5x + \cos4x} {1-2\cos3x} = -\cos2x -\cos x $$ 
I tried using the obvious approach by converting $5x , 4x $ and $ 3x$ to either $2x$ or $x$ but all that seemed to do was to further complicate the fraction. Any hints would be much appreciated.
I also tried applying various identities but to no avail.

Main idea: Stick to the $\cos x$ function and use brute force.
LHS:
The numerator
$=\cos5x+\cos4x$
$=\cos3x\cos2x-\sin3x\sin2x+2\cos^22x-1$
$=(\cos2x\cos x-\sin2x\sin x)(2\cos^2x-1)-(\sin2x\cos x+\sin x\cos2x)(2\sin x\cos x)+2(2\cos^2x-1)^2-1$
$=((2\cos^2x-1)\cos x-2\sin^2 x\cos x)(2\cos^2x-1)-(2\sin x\cos^2x+\sin x(2\cos^2x-1))(2\sin x\cos x)+2(4\cos^4x-4\cos^2x+1)-1$
$=(2\cos^3 x-\cos x-2\sin^2x\cos x)(2\cos^2x-1)-(4\sin x\cos^2x-\sin x)(2\sin x\cos x)+8\cos^4x-8\cos^2x+1$
$=(2\cos^3 x-\cos x-2(1-\cos^2x)\cos x)(2\cos^2x-1)-(4(1-\cos^2x)\cos^2x-(1-\cos^2 x))2\cos x+8\cos^4x-8\cos^2x+1$
$=(4\cos^3x-3\cos x)(2\cos^2x-1)-(5\cos^2x-4\cos^4x-1)2\cos x+8\cos^4x-8\cos^2x+1$
$=8\cos^5x-6\cos^3x-4\cos^3x+3\cos x-10\cos^3x+8\cos^5x+2\cos x+8\cos^4x-8\cos^2{x}+1$
$=16\cos^5x+8\cos^4x-20\cos^3x-8\cos^2x+5\cos x+1$
$=(1+6\cos x - 8\cos^3 x)(1-\cos x - 2\cos^2 x)$
The denominator $=1-2\cos3x$
$=1-2(\cos2x\cos x - \sin2x\sin x)$
$=1-2((2\cos^2x-1)\cos x-2\sin^2x\cos x)$
$=1-2(2\cos^3x-\cos x-2(1-\cos^2x)\cos x)$
$=1-2(4\cos^3x-3\cos x)$
$=1+6\cos x-8\cos^3x$
The fraction $=1-\cos x - 2\cos^2 x$
$= - (2\cos^2x-1)-\cos x$
$=-\cos2x-\cos x$
$=$ RHS
