Trigonometric polynom Prove that $$\cos\frac{\pi}{7},\cos\frac{3\pi}{7},\cos\frac{5\pi}{7}$$ roots of polynomial $8x^3-4x^2-4x+1=0$
I'm confused, what can i do with $\frac{\pi}{7}$

If $\sin4x=\sin3x,$
$4\sin x\cos x\cos2x=3\sin x-4\sin^3x$
If $\sin x\ne0,$  $$4\cos x\cos2x=3-4\sin^2x\iff4\cos x(2\cos^2x-1)=3-4(1-\cos^2x)$$
$$\iff8\cos^3x-4\cos^2x-4\cos x+1=0$$
Again if $\sin4x=\sin3x,$
$4x=n\pi+(-1)^n3x$ where $n$ is any integer
For even $n=2m$(say), $x=2m\pi\implies \sin x=?$ where $m$ is any integer
For odd $n=2m+1$(say),  $7x=(2m+1)\pi$ where $m$ is any integer
$\implies x=\dfrac{(2m+1)\pi}7$ where $m\equiv0,\pm1,\pm2,\pm3\pmod7$
But for $m\equiv3\pmod7,\sin x=0$
So, the roots of $$8t^3-4t^2-4t+1=0$$ are $\cos\dfrac{(2m+1)\pi}7$ where $m\equiv0,1,2\pmod7$ 
as $\cos(-y)=\cos y,\cos\dfrac\pi7=\cos\dfrac{\pi\{2(-1)+1\}}7$ etc.
