<s>[INST] 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 $\frac{\pi}{7}$
 [/INST] Let $\theta=\pi/7$. Note that 
$$\cos(4\theta)=-\cos(3\theta).\tag{1}$$ 
Let $x=\cos\theta$. By the double-angle identity for cosine, we have $\cos(2\theta)=2x^2-1$, so $$\cos(4\theta)=2(2x^2-1)^2-1.$$ 
By the triple-angle identity for cosine we have
$$\cos(3\theta)=4x^3-3x.$$
Now using Equation (1) we obtain 
$$2(2x^2-1)^2-1=-(4x^3-3x).$$
This is a quartic equation. However, it has the obvious root $x=-1$. Divide  the polynomial $2(2x^2-1)^2-1+4x^3-3x$ by $x+1$, and we obtain the cubic of the question.
To finish, verify that if $\varphi=3\pi/7$ or $5\pi/7$, then $\cos(4\phi)=-3\cos(3\phi)$.
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