Solving $f(f(f(x)))=x$ where $f(x)=4x(1-x)$

Find the number of solutions of $f^{(3)}(x)=x$ where $f:[0,1]\to[0,1], \ f(x)=4x(1-x)$

I tried substituting $x=\sin^2(\theta)$ but it didn't help.

There is a post on the same question but I don't seem to grasp it.

The substitution $x = \sin^2 \theta$ with $\theta \in [0,\pi/2]$ does work:

$f(x) = 4x(1-x) = 4 \sin^2\theta(1-\sin^2\theta) = (2 \sin \theta \cos \theta)^2 = \sin^2(2\theta)$

That means that $f^{(3)}(x) = \sin^2(8\theta)$.

So you want to solve $\sin^2(8\theta) = \sin^2\theta$. Then

$$0=\sin^2(8\theta) - \sin^2\theta = (\sin(8\theta)+\sin\theta)(\sin(8\theta)-\sin\theta) $$ $$= 2 \sin(\tfrac{8\theta + \theta}{2})\cos(\tfrac{8\theta - \theta}{2})\cdot 2 \sin(\tfrac{8\theta - \theta}{2})\cos(\tfrac{8\theta + \theta}{2})=4\sin(\tfrac{9}{2}\theta)\cos(\tfrac{7}{2}\theta)\sin(\tfrac{7}{2}\theta)\cos(\tfrac{9}{2}\theta)$$

But

$\sin(\tfrac{9}{2}\theta)=0$ for $\theta = 0,\tfrac{2}{9}\pi,\tfrac{4}{9}\pi$

$\cos(\tfrac{7}{2}\theta)=0$ for $\theta = \tfrac{1}{7}\pi,\tfrac{3}{7}\pi$

$\sin(\tfrac{7}{2}\theta)=0$ for $\theta = 0,\tfrac{2}{7}\pi$

$\cos(\tfrac{9}{2}\theta)=0$ for $\theta = \tfrac{1}{9}\pi,\tfrac{1}{3}\pi$

So the answer are $x = \sin^2(\theta)$ for $\theta \in\{ 0,\tfrac{1}{9}\pi,\tfrac{2}{9}\pi,\tfrac{1}{3}\pi,\tfrac{4}{9}\pi,\tfrac{1}{7}\pi,\tfrac{2}{7}\pi,\tfrac{3}{7}\pi\}$