59th Ukrainian National Mathematical Olympiad

The equation can be written the following way: $$\frac{\sqrt{x+2}}{\sqrt{x+1}}=\frac{\cos 2x+3}{\cos 2x+1}.$$ Consider two functions: $f(t)=\frac{t+2}{t+1}$, $t\ge 0$ and $g(y)=\frac{y+3}{y+1}$, $y\in (-1;1]$.

If $t\ge 0$, then $f(t)=\frac{t+2}{t+1}<2\iff t+2<2t+2\iff t>0$. Also, $f(t)=\frac{t+2}{t+1}=2$ only if $t=0$.

If $y\in (-1;1]$, then $g(y)=\frac{y+3}{y+1}>2\iff y+3>2y+2\iff y<1$. Also, $g(1)=\frac{1+3}{1+1}=2$ only if $y=1$.

Thus, given equation holds only if $f(\sqrt{x})=2=g(\cos 2x)$, that is possible only if $\sqrt{x}=0$ and $\cos 2x=1$. Therefore, the only solution is $x=0$.