cos x = 1/x has infinitely many solutions

Idea. One sign change gives one root. To get infinitely many roots, find one in each of infinitely many disjoint intervals.

Setup. For $x>0$ the equation is equivalent to $\varphi(x)=x\cos x-1=0$. The function $\varphi$ is continuous.

Sign change. For an integer $k\ge1$:
- $\varphi(2k\pi)=2k\pi-1>0$,
- $\varphi((2k+1)\pi)=-(2k+1)\pi-1<0$.

By the intermediate value theorem there is $x_k\in(2k\pi,(2k+1)\pi)$ with $\varphi(x_k)=0$.

Conclusion. The intervals $(2k\pi,(2k+1)\pi)$ are pairwise disjoint, so the $x_k$ are pairwise distinct. The equation has infinitely many solutions.

Picture. The curve $y=\cos x$ keeps crossing the decreasing curve $y=\frac1x$, which becomes flat near $0$ while the cosine goes on oscillating between $-1$ and $1$.