Find number of solutions of $x = a\sin x$ How would you analytically find the number of solutions of the function $x = a\sin(x)$ with respect to $a$?
 A: Here $x=0$ is always a root. For other roots graphical solution is possible:
Re-write the equation as $\sin x=\frac{x}{a}=y$, then $y=x/a$ is a line whose slope is $1/a$
we know that $y=x$ is tangent to $y=\sin x$ so if $1/|a|<1 \implies |a|>1$ this equation will have root(s) and only one  root if $|a|<1.$ For instance $x=10 \sin x$ has many as $y=x/10$ will cit $y=\sin x$ curve.   but $x=0.5 \sin x$ has only one root $x=0$ because $y=2x$ dose not cut the curve $y=\sin x$. 
See the Fig below the blue line $y=x/10 (a=10)$ is cutting the $\sin x$ curve at several points but $y=2x (a=1/2)$ (red) line is meeting the $\sin x$ curve only at $x=0$.

A: The number of roots can actually be computed. First, exclude the trivial case of one root $x=0$ for $|a|\le 1$. Given the symmetry, consider only the positive roots below and for $a>1$. 
Locate the tangential points $x_n$ between the line $y=x$ and $y=a\sin x$, i.e. with matching functional and derivative values, $x_n = a\sin x_n$ and $1 = a\cos x_n$, which leads to solving the equation 
$$x_n = \tan x_n$$
Despite lacking analytic solutions, a fairly accurate algebraic approximation is known,
$$x_n=\frac{(1+4n)\pi}2 - \frac{2}{(1+4n)\pi}\tag 1$$
Next, identify the largest $n$ such that 
$$\frac1a < \frac{\sin x_n}{x_n} = \cos x_n=\sin\frac{2}{(1+4n)\pi}$$
or,
$$n<\frac 1{2\pi\csc^{-1}a} -\frac14\tag 2$$
Then, the number of positive roots is $2n+1$. Take the example of $a = 15$, the inequality (2) yields $n< 2.13$. Thus, the largest $n$ is 2 and the number of positive roots is 5. Note that the total number of roots, including negative ones and $x=0$, is then $4n+3$ for $a>1$. Similar analysis can be performed for the case of $a<-1$.
