# How many real numbers satisfy the following

How many real numbers satisfy: $$\sin x=\frac{x}{100}$$

I don't know where to start it how to do this at all. Can someone please help me?

• try things like $\sin x = \frac{x}{3}$ first, where you can draw the graphs of $y = \sin x$ and $y = \frac{x}{3}$ yourself. graph paper, five lines per inch: printablepaper.net/preview/grid-portrait-letter-5-index Valuable skill, drawing graphs – Will Jagy May 26 '15 at 0:54
• $5 \pi/2 \approx 7.85,$ so doing $\sin x = \frac{x}{8}$ for $-8 \leq x \leq 8$ should give a good idea of what is going on. – Will Jagy May 26 '15 at 1:38

Hint:

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

Solution:

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 $31+31+1 = 63$.