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.
 A: Swatting this fly with an oversized mallet.
Let
$$f(x)=\cos^5x+\cos^5\left( x+\frac{2\pi}{3}\right) + \cos^5\left( x+\frac{4\pi}{3}\right).$$
Because cosine has period $2\pi$, we easily see that $f(x)$ has period $2\pi/3$. It is well behaved (continuously differentiable), so it can be expanded into a Fourier series.
The function $f(x)$ is even, $f(-x)=f(x)$ for all $x$, so the expansion only contains cosine terms.
As you observed, it is a quintic polynomial in $\cos x$, so
$$
f(x)=\sum_{k=0}^5a_k\cos kx
$$
for some constants $a_k\in\Bbb{R}$. Period $2\pi/3$ implies that $a_k=0$ unless $k$ is divisible by three. At this point we know that
$$f(x)=a_0+a_3\cos 3x$$
for some constants $a_0,a_3$. We have
$f(x+\pi)=-f(x)$ for all $x$ implying that $a_0=0$ (you can see this also by observing that only odd powers of $\cos x$ occur in that quintic, or by observing that $\int_0^{2\pi}f(x)\,dx=0$).
Last but not least
$$a_3=f(0)=1+(-1/2)^5+(-1/2)^5=15/16.$$
So
$$f(x)=\frac{15}{16}\cos 3x.$$
A: Use $\cos (A+B)$ formula to show for a natural number $n$,
$$\cos nx + \cos \left(nx+\frac{2n\pi}{3}\right) + \cos \left(nx+\frac{4n\pi}{3}\right) = \begin{cases} 0 , \quad \text{when n is coprime to 3} \\ 3\cos nx , \quad \text{when n is a multiple of 3} \end{cases}$$
Combining this with the identity from other answer,
$$\cos^5 \theta = \frac{1}{16}\cos 5\theta + \frac{5}{16}\cos  3\theta + \frac{10}{16}\cos  \theta$$
it is easy to see that first and third terms, $n=1,5$, should sum to zero, while middle term remains to give
$$\cos^5x+\cos^5\left( x+\frac{2\pi}{3}\right) + \cos^5\left( x+\frac{4\pi}{3}\right) = \frac{15}{16}\cos  3x$$
A: Hint: $$\cos^5\theta =\frac{ 10 \cos(θ) + 5 \cos(3 θ) + \cos(5 θ)}{16}$$because  $$\cos^5 x=1/32{(e^{ix}+e^{-ix})}^5=\frac{e^{5ix}+e^{-5ix}+5(e^{3ix}+e^{-3ix})+10(e^{ix}+e^{-ix})}{32}$$
Note that there will be infinitely many solutions as pointed out in comments
A: 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 $p^3+q^3+r^3, p^2+q^2+r^2$
