Solve $\sin(ax) / \sin(x) = a/2$ for $x$ I am currently trying to solve $\sin(ax)/\sin(x) = a/2$ for $x$, where $x$ is between $0$ and $\pi$ and $a$ is a constant.
As I have limited skills in math, I cannot seem to solve this problem without a brute force solution using Matlab to calculate a numerical value for this equation, I was wondering if there is a mathematical solution to this.
$a$ is non-zero, and around $500,000$ for my application.
Any help would be appreciated, thank you.
 A: You always have the trivial solution $x=0$.
The next root is a small value (on the order of $1/a$), so that the sine is very well approximated by $x$ (the next term being $-x^3/6$). The equation is rewritten
$$\frac{\sin ax}{a\sin x}\approx\frac{\sin ax}{ax}=\text{sinc }ax=\frac12,$$
hence
$$x\approx\frac1a\text{sinc}^{-1}\frac12=\frac{1.895494267034\cdots}a$$
For large $a$ this initial approximation is excellent, if not just sufficient.
Then no other root is found until the end of the range, very close to $\pi$. Making the change of variable $\dfrac ta=\pi-x$, we can approximate the sine as $\dfrac ta$. The equation becomes
$$\sin(a\pi-t)\approx\frac t2.$$
By the periodicity of the sine, we can replace $a\pi$ by $b=a\pi\bmod 2\pi$, a more manageable coefficient.
Anyway, uness $a$ is integer, there is no further simplification here and the equation has the general form of the intersection of a straight line with a sinusoid, $\sin p=\alpha p+\beta$, that can have up to three solutions and needs to be solved numerically.
To get started, the two arches of the sinusoid can be approximated by $4p(\pi-p)/\pi^2$ and $-4(p-\pi)(2\pi-p)/\pi^2$, leading to quadratic equations.
UPDATE: numerical issue
Actually, finding the root(s) close to $\pi$ is an ill-posed problem. Indeed, its location depends on the value of $a\pi\bmod2\pi$ so that $a$ needs to be known with, say $2$ exact decimals ($8$ significant digits). If this is not the case, all you can say is that $|\dfrac t2|<1$ and $x$ is in the range $(\pi-2/a,\pi)$.
A: Let
$y = ax$,
so
$\sin(ax)/\sin(x) = a/2$
becomes
$\sin(y)/\sin(y/a) = a/2$.
Since $a$ is large,
under the assumption that
$y/a$ is small,
I will approximate 
$\sin(y/a)$
by
$y/a$,
so this becomes
$a \sin(y)/y = a/2$
or
$\sin(y)/y = 1/2$.
The solution to this,
according to alphy,
is
$y \approx ±1.89549$,
so that
$x = y/a
\approx ±1.89549/a
$.
