Is there an easy way to see that $f_n(x) = x/n - \cos x$ for all $x\in \mathbb{R}$ has finitely many zeros for each $n$?
I was tutoring someone in calculus and one of the questions was to show that a certain function has a unique zero. The question got me wondering about similar questions and I can't seem to prove given statement.
as PhoemueX says in the comments, a zero must satisfy $|x| \le n$. To finish the argument from here, by the mean value theorem, between any two zeroes $r, s$ there must be a point $a \in (r, s)$ such that
$$f_n'(a) = \frac{1}{n} - \sin a = 0$$
hence such that $\sin a = \frac{1}{n}$, and this occurs at most twice each period. So there are at most $3$ zeroes in each period (actually there are at most $2$ zeroes but we only need any finite bound), and there are a bounded number of periods in the interval $[-n, n]$, so we're done.
More generally, a strictly convex twice-differentiable function (satisfying $f''(x) > 0$) intersects a line at most twice by the same argument, because $f'(x)$ is strictly increasing and so takes on any particular value at most once, and $\cos x$ alternates between being convex and concave a finite number of times in each period. Conversely, a continuously twice-differentiable function with infinitely many isolated zeroes in an interval must have the property that $f''(x)$ changes sign infinitely often.
This is more or less a formalization of the intuition you get from staring at the graph: