Consider the following transcendental equation:
$$\phi = \beta \sin \phi . \qquad (*)$$
How does one generate a description of how $\phi$ depends on $\beta$?
My attempt
From inspection (i.e. drawing pictures) one finds that for $\beta<1$ there is a single solution: $\phi=0$. At $\beta=1$ the solution at $\phi=0$, $\beta=0$ splits into three solutions, and then as $\beta$ increases additional solutions continue to appear in pairs.
The only technique I know is to assume that $\phi$ is implicitly given as a function of $\beta$. One then writes this function as a Taylor series and solves for the coefficients up to desired order.
Supposing $\beta$ is a function of $\phi$ we write a Taylor series $$\beta(\phi) = a\phi + b\phi^2 + \cdots.$$ Using this and expanding the $\sin$ function in $(*)$ we find
$$ \begin{align} \phi &= (a\phi + b\phi^2 + \cdots)(\phi - \phi^3/6 \cdots) \\ \phi &= a\phi^2 + b\phi^3 - \frac{a}{6}\phi^4 - \frac{b}{6}\phi^5 + \cdots \end{align} $$
There are no solutions other than $\phi=0$, $\beta=0$, so we have not captured the new pair of solutions which appears as $\beta$ increases past 1. This is strange. The Taylor series could have failed because $d\beta/d\phi$ is diverging at our expansion point. To check this, differentiate $(*)$ with respect to $\phi$ to find $$\frac{d\beta}{d\phi} = \frac{1 - \beta(\phi) \cos\phi}{\sin\phi}.$$ As $\phi$ approaches 0 from the right, we have $d\beta/d\phi = (1-\beta(\phi))/\phi$. This may or may not diverge; it depends on the functional form of $\beta(\phi)$.
The same ambiguity shows up if one instead regards $\phi$ as an implicit function of $\beta$.
From a numerical solution I can see that in fact $d\phi/d\beta = \infty$ at $\beta=1$. That suggests that the failure of the Taylor series for $\beta$ in terms of $\phi$ was not due to diverging derivative. This suggests that $\beta$ cannot be written as a Taylor series in $\phi$ for a different reason, e.g. fractional powers are needed. How then, does one proceed?
