Solutions for trigonometric equation

Question

Consider the following equation:

$$ 2 = \cos(x) + \cos(\sqrt{2}x) $$

So, I'm reading this proof which says that the equation has a solution iff

$$ x=0 \mod{2\pi} \quad \wedge \quad \sqrt{x}=0 \mod {2\pi} $$

Now, why is it the only option?

Answer 1

As for real $y,\cos y\le1$

We need $\displaystyle\cos x=\cos(\sqrt2x)=1$

$\displaystyle\cos x=1\implies x=2m\pi $ where $m$ is any integer

and $\sqrt2x=2n\pi$ where $n$ is any integer

If $mn\ne0, \frac x{\sqrt2x}=\frac mn\iff \frac mn=\frac1{\sqrt2}$ which is impossible

Hence, $m,n$ must be zero