With Lagrange reversion:
$$x\sin(x)=1\iff x=\{2\pi k+\csc^{-1}(x),(2k+1)\pi-\csc^{-1}(x)\}$$
Therefore:
$$x_{2k}=2k\pi+\sum_{n=1}^\infty\frac1{n!} \left.\frac{d^{n-1}}{dw^{n-1}}\csc^{-1}(w)^n\right|_{2k\pi} \\x_{2k+1}=(2k+1)\pi+\sum_{n=1}^\infty\frac{(-1)^n}{n!}\left.\frac{d^{n-1}}{dw^{n-1}}\csc^{-1}(w)^n\right|_{(2k+1)\pi}$$
with examples shown here. Next use Stirling S1:
$$\begin{align}\frac1{n!}\frac{d^{n-1}}{dw^{n-1}}\csc^{-1}(w)^n=(-i)^n \frac{d^{n-1}}{dw^{n-1}}\frac{\ln^n\left(\frac{\sqrt{w^2-1}+i}w\right)}{n!}=(-1)^n\sum_{m=n}^\infty\frac{S_m^{(n)}}{m!}\frac{d^{n-1}}{dw^{n-1}}\left(\frac{\sqrt{w^2-1}+i}w-1\right)^m\end{align}$$
Also, the binomial theorem presents $\displaystyle\frac{d^{n-1}}{dw^{n-1}}\left(\frac{\sqrt{w^2-1}+i}w\right)^j$, but that is still too complicated. Additionally, no generating functions from these polynomials matched it.
What is $\displaystyle\frac{d^{n-1}}{dw^{n-1}}\csc^{-1}(w)^n$ as a triple sum or simpler? We could find all solutions to $\csc(x)=x$ explicitly then.