I need solve the funcional equation $$f(4x(1-x))=\sin(\pi f(x))\tag{1}$$
Remark: I have tried to solve this equation but I have only reached a reformulation: $$ \frac{4(1-2y)f'(4y(1-y))}{\sqrt{1-f^{2}(4y(y-1))}}=\pi f'(y) .$$
Edit: This question is motivated by a problem of Ergodic Theory, the problem is find invarian measure of dynamical system $$x_{n+1}=\sin(\pi x_{n}).\tag{2}$$ For this purpose it is sufficient to find its invariant density. In this sense, the idea is find a function $f$ inyective such thar satisfies (1) with $f(0)=0$ and $f(1)=1$. If that function existed then we have a change of coordinates given by $$x=f(y).$$ Therefore, sustituting in (2) we have $f(y_{n+1})=\sin(\pi f(y_{n}))$, but by (1) we have $$ f(y_{n+1})=f(4 y_{n}(1-y_{n})) .$$ Since we assume that $f$ injective then $$y_{n+1}=4y_{n}(1-y_{n}) \tag{3}.$$ But we know the invariant probability density function of dynamical system (3) is the function $$\rho(x)=\frac{1}{4\sqrt{x(1-x)}}.$$ Therefore, the invariant probability density function g of dynamical system (1) is the function $$g(x)=\left|\frac{df(x)}{dx}\right|\rho(x).$$ Therefore, the whole problem is reduced to finding $f$.