FINAL ROUND

Assume that there exists a function $f(x)$ satisfying the problem condition: $$\{\begin{array}{l}\{f(x)\}{\sin }^{2}x+\{x\}\cos f(x)\cos x=f(x),\\ f(f(x))=f(x),\end{array}$$ for all real $x$. Replacing $x$ by $f(x)$ in the first equality, we obtain $$\{f(f(x))\}{\sin }^{2}f(x)+\{f(x)\}{\cos }^{2}f(x)=f(f(x)).$$ Since $f(f(x))=f(x)$, from the obtained equality it follows that $\{f(x)\}=f(x)$. So $f:\mathbb{R}\to [0;1]$. Replacing $x$ by $\pi$, we have $-\{\pi \}\cos f(\pi )=f(\pi )$. Since $f(\pi )\in [0,1]$ and $\{\pi \}\ne 0$, we see that the left-hand side of the last equality is negative, whereas the right-hand side is nonnegative, a contradiction.