<s>[INST] Number of solutions of $\cos^5x+\cos^5\left( x+\frac{2\pi}{3}\right) + \cos^5\left( x+\frac{4\pi}{3}\right) =0$ 
Solve in the interval $[0,2\pi]$ : $$\cos^5x+\cos^5\left( x+\frac{2\pi}{3}\right) + \cos^5\left( x+\frac{4\pi}{3}\right)=0 $$

I tried expanding the L.H.S by applying the formula of $\cos(A+B)$ but it results in a quintic polynomial in terms of $\cos x $. Wolfram alpha has simplified the left hand side all the way to $\frac{15}{16}\cos3x$ but I'm unable to think of any method to reach there.
 [/INST] Clearly the roots of $$4c^3-3c-\cos3x=0$$ are $c_r\cos(x+2r\pi/3), r=0,1,2$
Let $p=c_0,c_1=q,c_2=r$
$p+q+r=0\implies p^3+q^3+r^3=3pqr$
$pq+qr+rp=-3/4\implies p^2+q^2+r^2=0^2-2(-3/4)$
$pqr=\dfrac{\cos3x}4$
As $4c^5=3c^3+c^2\cos3x,$
$4(p^5+q^5+r^5)$
$=3(p^3+q^3+r^3)+(p^2+q^2+r^2)\cos3x$
Replace the values of $p^3+q^3+r^3, p^2+q^2+r^2$
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