55rd Ukrainian National Mathematical Olympiad - Fourth Round

Transform the equation to the form: $$\cos [x]\cdot \cos \{x\}=\sin [x]\cdot \sin \{x\},$$ if $\sin [x]\cdot \sin \{x\}\ne 0$. Then we have $$\cos [x]\cdot \cos \{x\}-\sin [x]\cdot \sin \{x\}=\cos ([x]+\{x\})=\cos x=0.$$ Hence $x=\frac{1}{2}\pi +\pi k,k\in \mathbb{Z}$. Remains that all these values satisfy the initial condition. Indeed, none of them are integer, hence $\{x\}\ne 0\Rightarrow \{x\}\in (0;1)\Rightarrow \sin \{x\}\ne 0$. Also since $\sin y=0$ implies $y=\pi n,n\in \mathbb{Z}$, hence the expression is equal to zero for integer $y$ only when $y=0$. Notice that for $k\ge 0$ $x=\frac{1}{2}\pi +\pi k>1$ and $[x]>0$, and for $k<0$ $x=\frac{1}{2}\pi +\pi k<-1$ and $[x]<0$. Hence all $x$ that were found satisfy the equation.