Inverse of $f(x)=\sin(x)+x$ What is the inverse of 
$$f(x)=\sin(x)+x.$$
I thought about it for a while but I couldn't figure it out and I couldn't find the answer on the internet.
What about
$$f(x)=\sin(a \cdot x)+x$$
where $a$ is a known real constant.
Thank you for taking the time to read this question!
Sorry if this has been asked before...

Why not give another “closed form” since @GEdgar’s solution for the Kepler equation technically has $E(a,M)$ as the solution in the Wikipedia link. Here is a solution to the inverse of $y=\sin(x)+x$ using mathematica function using Inverse Beta Regularized which is a standard function introduced in $1996$. The answer is from:

Closed form of $x$ for $x=\cos(x)$: Intuition for why the Dottie number is an inverse sine of the median of a Beta distribution.

where

$$M=E-e\sin(E)\iff x=y-a\sin(y)\implies y=\text E(a,x)$$

and

$$\text E(-1,x)=2\sin^{-1}\sqrt{\text I^{-1}_{\frac x\pi}\left(\frac12,\frac32\right)}=\text{hav}^{-1}\left(\text I^{-1}_{\frac x\pi}\left(\frac12,\frac32\right)\right)$$

Here is a numerical tester for the inverse function
Here is a plot of:
$$\boxed{x=\sin(y)+y\implies y=\text E(-1,x) = \text{hav}^{-1}\left(\text I^{-1}_{\frac x\pi}\left(\frac12,\frac32\right)\right),0\le x\le \pi}: $$

The code is inversehaversine(inversebetaregularized(x/pi,1/2,3/2)) 
where appears the Inverse Haversine, a transformed inverse sine function.
The differential equation for the inverse of $y=\sin(x)+x$ is:
$$y’\text{vercos}(y))=1$$
where vercos is the versed cosine, a transformed cosine function. The inverse function’s domain is extended by using a series expansion of $\text I^{-1}_z(a,b)$. Please correct me and give me feedback!
Unfortunately, it seems that no more solutions of the inverse of $x-a\sin(x)=y$ can be solved for using this method. The only exception is that the inverse of $x-\sin(x)$ can be given in closed form by switching the $\frac32$ and
$\frac12$ in the boxed answer.
