For how many values of $x$ does $$f(x)=\cos x+\cos(\sqrt{2}x)$$ attain its absolute minimum?
My Attempt:
Absolute maximum of $f(x)$ is clearly $2$ which occurs when $x=0$. But what should be the approach to obtain absolute minimum.
One can observe that $f(x)$ is bounded, continuous, even and non-periodic so it will achieve its absolute minimum at least twice.
Can we say that it will achieve its minimum(absolute) exactly twice. And if yes what will that minimum be and for what value(s) of $x$ will it occur
Considering that cosine is even was a noble effort, but this problem was looking for something different.
Since the minimum of $\cos$ is -1, $f$ cannot attain a value less than $-2$. Can it attain that value? If it did, there would have to be some real $x$ such that $x$ and $x\sqrt2$ were odd multiples of $\pi$, but that is impossible since $\sqrt2$ is irrational.
However, we can find odd integers $x$ such that $x\sqrt2$ is as close to an odd integer as we please. (As the comments point out, this is due to the fact that the fractional parts of $x\sqrt2$ are dense in the unit interval, and just trust us if you don't understand what that means.) Therefore, the range of $f$ is $(-2,2]$, and it has no global minimum at all.