https://artofproblemsolving.com/community/c6h1220090p6096103

parmenides51 wrote: Indicate (justifying your answer) if there exists a function $f: R \to R$ such that for all $x \in R$ fulfills that i) $\{f(x))\} \sin^2 x + \{x\} cos (f(x)) cosx =f (x)$ ii) $f (f(x)) = f(x)$ Using ii) and plugging $x\to f(x)$ in i), we get $\{f(x)\}=f(x)$ and so $f(x)\in[0,1)$ Plugging this in i), we get $\{x\}\cos f(x)\cos x=f(x)\cos^2x$ and so $\{x\}\cos f(x)=f(x)\cos x$ $\forall x\ne \frac{\pi}2+k\pi$ But, setting $x\to\frac{\pi}2$ in above equality, $RHS\to 0$ (since $f(x)\in[0,1)$ while LHS is the product of something tending towards $\frac{\pi}2-1$ and something $\in(\cos 1,1]$ and so can never approach $0$ And so $\boxed{\text{No such function}}$