Obtaining a closed form solution for $\frac{x}{\sin(x)} = a$ on $[0,1)$

Question

Is it possible to obtain a closed form expression for the root of $$\frac{x}{\sin(x)} = a,$$ where the constant $a \in \left[1,\frac{1}{\sin(1)}\right]$ and $x \in [0,1)$?

Answer 1

We can rewrite the equation as

$$x = a\sin x=\Phi(x).$$

If $\Phi(x)$ is fulfilling the conditions for the fixed-point-theorem we can state that the solution is given by

$$x=a\sin(a\sin(a\sin(a\sin(...))).$$

Answer 2

Consider the function

$$f(x)=x−a\sin(x).$$

Differentiating gives

$$f′(x)=1−a\cos(x) \hspace{2mm}\text{ and }\hspace{2mm} f′′(x)=a\sin(x)\gt0 \forall x\in[0;1)$$

then

$$f′([0;1))=[1−a; 1−\cos(1))$$

which means that there is a unique $\alpha$ such that $f'(\alpha)=0$. The minimum of $f$ is therefore $f(\alpha)$. Since

$$f(1) = 1 − a\sin(1)\gt0 \hspace{2mm}\text{ and }\hspace{2mm} f(0)=0$$

then $f(\alpha)\lt0$ and $f$ has two roots $0$ and $\beta$. Thus there is no closed form for $\beta$, only a numerical approach.

Answer 3

Too long for a comment

We know that $$\frac x{\sin(x)}=1+\sum_{n=1}^\infty \frac{\left(4^n-2\right) \left| B_{2 n}\right| }{(2 n)!}\,x^{2n}$$ So, all coefficients being known, we can use power series reversion and write $$x=\sum_{n=0}^\infty (-1)^{n}\, \alpha_n\,t^{2n+1}\quad \text{where} \qquad t= \sqrt{6(a-1)}$$

Since we know the coefficients of the initial series, using the explicit formula given for the $n^{\text{th}}$ term by Morse and Feshbach,we know all the $\alpha_n$.