How to solve $\cos(2^n x) = \cos(x)$

Question

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

Answer 1

With only basic results on trigonometric equations: \begin{align}\cos(2^nx)=\cos x &\iff 2^nx\equiv \pm x\mod 2\pi\iff\begin{cases}(2^n-1)x\equiv 0\mod 2\pi\\(2^n+1)x\equiv 0\mod 2\pi\end{cases}\\[1ex] &\iff x\equiv 0\mod \frac{2\pi}{2^n-1},\bmod \frac{2\pi}{2^n+1}. \end{align}

Answer 2

Using the identity $$\cos A - \cos B = 2\sin\left(\dfrac{B-A}{2}\right)\sin\left(\dfrac{B+A}{2}\right),$$ we have $$0 = \cos(x)-\cos(2^nx) = 2\sin\left(\dfrac{2^n-1}{2}x\right)\sin\left(\dfrac{2^n+1}{2}x\right),$$ which holds iff $$\sin\left(\dfrac{2^n-1}{2}x\right) = 0 \ \ \text{OR} \ \ \sin\left(\dfrac{2^n+1}{2}x\right) = 0$$ $$\dfrac{2^n-1}{2}x = k\pi \ \ \text{OR} \ \ \dfrac{2^n+1}{2}x = k\pi$$ $$x = \dfrac{2k\pi}{2^n-1} \ \ \text{OR} \ \ x = \dfrac{2k\pi}{2^n+1}$$ for some $k \in \mathbb{Z}$.