How to solve cos⁡(2nx)=cos⁡(x) I am given the function F:R→R , x↦4x(1−x) and I have...

How to solve ${\displaystyle \cos \left(2^{n}x\right)=\cos \left(x\right)}$
I am given the function ${\displaystyle F:\mathbb{R}\to \mathbb{R}}$ , ${\displaystyle x\mapsto 4x(1-x)}$ and I have to find all n-cycles of the function. I have already reduced this to the problem of solving ${\displaystyle \cos \left(2^{n}x\right)=\cos \left(x\right)}$ for ${\displaystyle x\in [0,\pi ]}$. Using Mathematica and ${\displaystyle n\in \{1,2,3,4\}}$, I concluded that the solutions must be ${\displaystyle \{\frac{2k\pi }{2^n-1},\frac{2k\pi }{2^n+1}\}\cap [0,\pi ]}$. Since the equation obviously has ${\displaystyle 2^n}$ solutions and the solution set has ${\displaystyle 2^n}$ elements, it only remains to verify that they are indeed solutions to the equation. I tried using ${\displaystyle \cos \left(2x\right)=2{\cos }^2\left(x\right)-1}$ and proving this by induction on n, but I was not able to.