I'm trying to solve an exercise in which I need to prove that the logistic map with parameter $\mu = 4$, $F_4:[0,1]\to[0,1]$, $F_4(x) = 4x(1-x)$, satisfies that for every positive integer $k$, $F_4^k$ has $2^k$ fixed points.
My approach has been to prove that $F_4$ is conjugated to the degree 2 Chebyshev polynomial $q_2(y) = 2y^2-1$, for $y \in [-1,1]$, via the diffeomorphism $h:[0,1]\to[-1,1]$, $h(x) = 1-2x$.
Given that $F_4^k = h^{-1} \circ q_{2^k} \circ h$, my final result is \begin{equation} F_4^k(x) = \frac{1}{2} - \frac{1}{2} \cos\left [2^k \arccos{(1-2x)}\right] \end{equation}
However, I am not being able to solve the equation $ F_4^k(x) = x$ in order to find the fixed points, and I am not finding a proper method to prove that indeed $ F_4^k(x) = x$ has $2^k$ solutions for $x \in [0,1]$. Thanks in advance for your help.
Edit: Now, I have seen that computing the fixed points of $F_4^k(x)$ is equivalent to solving the equation \begin{equation} q_{2^k}(y) = y \end{equation} i.e., finding the fixed points of the $2^k$-th degree Chebyshev polynomials. By graphical analysis I have concluded that this equation has exactly $2^k$ solutions, but I don't know how to prove it.
Take a middle approach, use the parametrization behind Chebyshev and set $x=\sin^2\theta$, $θ\in[0,\frac\pi2]$. Then $$ F_4(\sin^2θ)=4\sin^2θ\cos^2θ=\sin^2(2θ) $$ so that also $$ F_4^k(\sin^2θ)=\sin^2(2^kθ) $$ On each segment $θ\in[2^{-k-1}\pi\,m,\,2^{-k-1}\pi\,(m+1)]$, $m=0,1,...,2^k-1$, the values of $F_4^k$ span the full interval $[0,1]$, guaranteeing an intersection with the diagonal via the intermediate value theorem.