During the investigation of my thesis I found the following problem:

I need find a injective function $f$ such that 
$$f(4x(1-x))=\sin(\pi f(x))\tag{1}$$
and  $f(0)=0$ and $f(1)=1$.

**Remark:** I have tried to solve this equation but I have only reached a reformulation:
$$
\frac{4(1-2y)f'(4y(1-y))}{\sqrt{1-f^{2}(4y(y-1))}}=\pi f'(y)
.$$

***You might think that this problem is not suitable for this site but for me it is important because it arises in trying to address the following situation.***

**The situation:**
 This question is motivated by a problem of Ergodic Theory in my thesis, the problem is find  invariant 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 $f$ injective such that 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, substituting 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 $f$.