Trigonometric equality $x = 99 \sin (\pi x)$ Find the number of real solutions of $\displaystyle x = 99 \sin (\pi x)$. 
I am getting stuck in some trigonometric relations.
 A: There is an obvious root at $x=0$. Since $\sin(-\pi x)=-\sin(\pi x)$, there are just as many negative solutions as there are positive solutions. 
So we count the positive solutions, double the result, and add $1$. Now we show how to count the positive solutions. We will use a picture to do the count. 
Rewrite the equation as
$$\sin(\pi x)=\frac{x}{99}.$$
For $x\gt 99$, the right-hand side is $\gt 1$, while the left is always $\le 1$. So we will not have to look for solutions beyond $x=99$. 
To count the positive solutions $\lt 99$, draw the graph of $y=\sin(\pi x)$. We need to go a moderately long way. Note that $\sin(\pi x)$ has period $2$. So when we go up to $x=99$, there will be $49$ full periods, plus a half period. 
Let's take a look at $y=\frac{x}{99}$. This is a line with very shallow slope. It enters the first period of $y=\sin(\pi x)$ and leaves it, giving $2$ intersections, one in the interval $(0,\pi/2)$, the other in the interval $(\pi/2,\pi)$. It enters and leaves the second period ($2$ more solutions). Continue. You will have to think a little about what happens in the final half-period. 
Remark: A graphing calculator may help, but you will need to play with the viewing window. 
