51st Ukrainian National Mathematical Olympiad, 4th Round

Answer: $x=\frac{3}{2}+2n,n\in \mathbb{Z}$.

Since $a-[a]=\{a\}$, where $\{a\}$ is the fractional part of $a$, we can rewrite our equation in the following way: $$\cos \pi x=\left[\frac{\{x\}}{2}-\frac{1}{2}\right].$$ Obviously, $0\le \frac{\{x\}}{2}<1$ for every real $x$. Consider two cases:

1) Let $0\le \frac{\{x\}}{2}<\frac{1}{2}$. Then $-\frac{1}{2}\le \frac{\{x\}}{2}-\frac{1}{2}<0$, and thus $[\frac{\{x\}}{2}-\frac{1}{2}]=-1$. So in this case we get the equation $\cos \pi x=-1$. The solutions of this equation are $x=1+2k,k\in \mathbb{Z}$. But for such $x$ we have that $\{\frac{x}{2}\}=\{\frac{1}{2}+k\}=\frac{1}{2}$, which contradicts our assumption. So, we obtain that there are no solutions in this case.

2) Let $\frac{1}{2}\le \frac{\{x\}}{2}<1$. Then $0\le \frac{\{x\}}{2}-\frac{1}{2}<\frac{1}{2}$, which implies that $[\frac{\{x\}}{2}-\frac{1}{2}]=0$. So, in this case our equation reduces to the equation $\cos \pi x=0$. The solutions for this equation are $x=\frac{1}{2}+k$, $k\in \mathbb{Z}$. For such $x$ we have: $$\{\begin{array}{l}\frac{x}{2}\\ \frac{1}{2}\end{array}=\{\begin{array}{l}\frac{1}{4}+k\\ \frac{1}{2}\end{array}=\{\begin{array}{ll}\frac{1}{4},& k=2n,\\ \frac{3}{4},& k=2n+1.\end{array}$$ So, for $\frac{1}{2}\le \frac{\{x\}}{2}<1$ we should take $k=1+2n$, $n\in \mathbb{Z}$. Therefore, $x=\frac{3}{2}+2n$, $n\in \mathbb{Z}$.