59th Ukrainian National Mathematical Olympiad

The equation can be written as: $$\frac{2\cos 2x}{\cos 2x+1}=\frac{6-3\cos 3x}{\cos 3x+2}.$$ Consider the following two functions: $f(x)=\frac{2x}{x+1}$ and $g(x)=\frac{6-3x}{x+2}$, $x\in (-1;1]$. If $x\in (-1;1)$, then $$f(x)=\frac{2x}{x+1}<1\iff 2x<x+1\iff x<1,g(x)=\frac{6-3x}{x+2}>1\iff 6-3x>x+2\iff x<1.$$

Thus, the equality holds only if $\cos 2x=\cos 3x=1$. A sketch of the graphs of the functions can lead to the same conclusion.

In order for the last equation to be true, the following has to hold: $2x=2\pi k$, and $3x=2\pi l$, $k,l\in \mathbb{Z}$. Due to the first equation $x\in \{-\pi ,0,\pi \}$, with the help of the second equation it is easy to check that the only answer is $x=0$.