proximo wrote: Find all functions $f(x)$ determined on interval $[0,1]$, satisfying following conditions $$[f(x)]\sin^{2}x+[x]\cos f(x)\cos x=f(x)$$$$f(f(x))=f(x)$$Here $[y]$ means a fractional part of number y Let $P(x)$ be the assertion $\{f(x)\}\sin^2x+\{x\}\cos f(x)\cos x=f(x)$, true $\forall x\in[0,1]$ $P(f(x))$ $\implies$ $\{f(x)\}=f(x)$ and so $f(x)\in[0,1)$ So $P(x)$ becomes $f(x)\sin^2x+\{x\}\cos f(x)\cos x=f(x)$ which is $\{x\}\cos f(x)\cos x=f(x)\cos^2x$ And since $\cos x\ne 0$ $\forall x\in[0,1]$, this is $\{x\}\cos f(x)=f(x)\cos x$ And so,$f(1)=0$ and $\forall x\in[0,1)$ : $\frac x{\cos x}=\frac{f(x)}{\cos f(x)}$ But $\frac x{\cos x}$ is injective (strictly increasing) over $[0,1]$ and so $f(x)=x$ $\forall x\in[0,1)$ Hence the answer : $\boxed{f(x)=\{x\}\text{ }\forall x\in[0,1]}$ which indeed is a solution

pco, what if $x \in R$? Is the solution the same?

I think this is wrong..please find mistake

solution