Sine vs. Line

The RHS takes on the values $-1 \le \sin{x} \le 1$ .

The LHS is within these values when $-1 \le \frac {x}{100} \le 1\quad \Rightarrow -100 \le x \le 100$ .

Within this range of $x$ , we note that $\frac {x}{100}$ straight line cuts the $\sin {x}$ curve at $2$ points (roots) in the positive half-cycle of the sine curve when $x>0$ or negative half cycle when $x<0$ .

Within $-100 \le x \le 0$ and $0 \le x \le 100$ , each has $\frac {100}{2\pi} \approx 15.9$ full cycles and $16$ positive or negative cycles.

Therefore the $\frac {x}{100}$ straight line cuts through $16\times 2 -1 = 31$ negative points and $31$ positive points and share $1$ point when $x=0$ .

Therefore the number of points cut and hence the number or real roots is $31+31+1 = \boxed{63}$ .

The following graph shows the $\sin{x}$ curve and $\frac {x}{100}$ straight line.

The graph

Let a solution be a. Since both x/100 and sin(x) are odd, it follows that -a is also a solution. If we draw, the 2 graphs, we see that there will be an intersection if sin(pi/2+2npi)=1>(pi/2+2npi)/100. Rearranging this gives us n<15.6.. so n=0,1,2,....,15. Each n gives 2 solutions.

However, n=0,-1,-2,-3,...-15 also give 2 solutions each. However, we have counted x=0 twice. So the answer is 4(16)-1=63.

$-1\leq \sin x\leq 1 \rightarrow -1\leq \frac{x}{100}\leq 1 \rightarrow -100\leq x\leq 100$

$100/{2\pi} \approx 15.9155$

For each period $\frac{x}{100}$ touches to $\sin x$ two times. But $0$ are counted two times so real roots are $16\times 2-1=\boxed{63}$