FINAL ROUND

Let $f(x)$ satisfies the problem condition: $$\{\begin{array}{l}\{f(x)\}{\sin }^{2}x+\{x\}\cos f(x)\cos x=f(x),\\ f(f(x))=f(x),\end{array}(1)(2)$$ for all real $x$. Replacing $x$ by $f(x)$ in (1) we obtain: $$\{f(f(x))\}{\sin }^{2}f(x)+\{f(x)\}{\cos }^{2}f(x)=f(f(x)).$$ From (2) it follows that $\{f(x)\}=f(x)$. So $f:[0,1]\to [0,1)$. From (1) and the equality $\{f(x)\}=f(x)$ it follows that: $$\{x\}\cos f(x)\cos x=f(x)(1-{\sin }^{2}x)=f(x){\cos }^{2}x.(3)$$ We have $\cos x>0$ and $\{x\}=x$ for $x\in [0,1]$. Therefore from (3) we have $x\cos f(x)=f(x)\cos x$ for $x\in [0,1]$. Dividing the last equality by $\cos x\cos f(x)$ we obtain: $$\frac{x}{\cos x}=\frac{f(x)}{\cos f(x)}.(4)$$ Consider the function $g(x)=x/\cos (x)$, $g:[0,1]\to \mathbb{R}$. It is increasing since $x$ increases and $\cos x$ decreases. Since in view of (4) $g(x)=g(f(x))$ for $x\in [0,1]$, we obtain $f(x)=x$.