Quadratic Recurrence $a_{n+1} = 4a_n - 4a_n^2$

Question

How do I solve this quadratic recurrence $a_{n+1} = 4a_n - 4a_n^2$?

Actually, it came from $f(x) = 4x(1-x)$ when $f : [0,1] \rightarrow [0,1]$.

The problem asked about number of distinct roots of $f^{1992}(x)$=x. I do know that there are $2^{1992}$ roots. But , I don’t know how to cancel the repeated roots.

I tried set $a_n = f(a_{n - 1})$ and $a_1$ = $c$.

Which gives the quadratic recurrence $a_{n+1} = 4a_n - 4a_n^2$. But , I don’t know how to solve it.

Edit : If this recurrence can’t be solved , I tried put $x = sin^2(x)$ and it yields $f(sin^2(x)) = sin^2(2x)$!

Edit 2: $f^n(x) = f(f^{n-1}(x))$.

Answer 1

Let $a_n=\sin^2(b_n)$ to make $$\cos (2 b_{n+1})=\cos (4 b_n)\implies b_n=2^{n-1} C\implies a_n=???$$