$\cos(x)+\cos(x\sqrt{2})$ is not periodic

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

Let $x=0$. Then

$$2=\cos T + \cos \sqrt{2}T$$

Since the cosine is at most one, this means that simultaneously $\cos T = 1$ and $\cos \sqrt{2} T = 1$. This is equivalent to:

$$T = 2\pi n, \ \mbox{for some } n \in \mathbb{Z}$$ $$\sqrt{2}T = 2\pi m, \ \mbox{for some } m \in \mathbb{Z}$$

Substitute $T$ from the first into the second:

$$\sqrt{2} (2\pi n)= 2\pi m$$ $$\sqrt{2}=\frac{m}{n}$$

So $\sqrt{2} \in \mathbb{Q}$ which is a contradiction, or $T=0$. Either way, you're done.

There's only one value of $x$ for which $f(x)=2$. Proof: Since $\cos\phi\le1$ this is only possible, if $\cos x = 1$ and $\cos(x\sqrt2) = 1$ hold true at the same time. The first statement is equivalent to $\frac{x}{\pi}\in\mathbb Z$, the second is equivalent to $\frac{x}{\pi}\sqrt2\in\mathbb Z$. The value $x=0$ fulfills the requirements. If a non-zero $x$ would fulfill both $\frac{x}{\pi}\in\mathbb Z$ and $\frac{x}{\pi}\sqrt2\in\mathbb Z$, the you could devide the second integer by the first. The result would be a rational number with value $\sqrt2$. This is a contradiction. Hence, only $x=0$ fulfills both requirements.