How many solutions has the equation $\sin x= \frac{x}{100}$ ?

First, we may suppose $x\ge 0$ since both sides are odd functions.

Using the Intermediate value theorem, there'll be two non-negative solutions on each interval $]2k\pi,2(k+1)\pi[$ as long as $\frac x{100}\le 1$, i.e. $x\le 100$. There results the number of non-negative solutions is equal to $2\times \biggl\lfloor \dfrac{50}{\pi}\biggr\rfloor=32$.

Hence, by symmetry, the total number of roots is $\;\color{red}{63}$.

Outside of the interval $[-100, 100]$ there are no solutions.

$x = 100\;$ occurs close to the right end of the interval $[30 \pi, 32 \pi]$

$x = -100\;$ occurs close to the left end of the interval $[-32 \pi, -30 \pi]$

Each interval $[2n\pi, (2n+2)\pi]$ will have 2 solutions for $n = -32$ to $n = 30$ except $x = 0$ will get counted twice.

So there are $2*32-1 = 63\;$ solutions.