This is my first post. I'm not a mathematician, just an electronics engineer who loves mathematics. In one of my projects, I arrived at the following function:

$$V\left(\varphi\right)=\frac{A\sqrt{\pi-\varphi+\sin{\varphi\cos{\varphi}}}}{\sqrt{2\pi}}$$

The project requires $V\left(\varphi\right)$ to be inverted, to obtain angle $\varphi$ (unknown), from a voltage $V$ (known). $V\left(\varphi\right)$ is continuous and strictly monotonic (descending), so an inverse mapping should exist. I tried to invert it symbolically, but couldn't arrive at a closed-form solution for $\varphi\left(V\right)$. I ended up using MATLAB to compute it numerically, and the project was successfully completed.

Out of pure curiosity, I asked my cousin (a mathematician) to attempt to symbolically invert the above function, but he also couldn't do it, and couldn't even give me a rigorous answer as to the existence of such solution. So, my questions are the following:

- Does a closed-form expression for $\varphi\left(V\right)$ exist?
- If the answer to (1) is YES, can someone provide that function, or point me to a method for deriving it?
- If the answer to (1) is NO, what is the formal reason for it? Is there a way to show/prove that such solution does not exist?

I apologise if this problem is too easy, too obvious, or even irrelevant to the MathOverflow community. I've already posted it in the Mathematics Stack Exchange community (for students and professionals), where I got some replies which loosely confirm my suspicion that a closed-form expression for $\varphi\left(V\right)$ does not exist. However, the replies were either too descriptive, or used Taylor series expansions to invert the function, which is not what I want. No reply provided rigorous answers to my questions. So, I thought of posting the problem here, where more advanced topics are discussed, in the hope that someone can provide me with some rigorous answers. This is not a homework exercise, and the associated practical problem has already been solved numerically. This post was made out of pure curiosity about the invertibility of functions of the form of $V\left(\varphi\right)$. Many thanks to all for your replies.

*Adding some graphics, in order to better illustrate the problem.*

The following figure shows how the function we're looking for, $\varphi\left(V\right)$, looks like. Notice that there appears to be no symmetry in this function. The values of $\varphi$ lie in the interval $[0,\pi]$, while the values of $V\left(\varphi\right)$ lie in the interval $[0,\frac {A} {\sqrt2}]$.

Based on the comment by @PietroMajer, the problem can be reduced to the inversion of function $k=x-\sin(x)$. In this case, the values of both $x$ and $k$ lie in the interval $[0,2\pi]$. The following figure shows a plot of the inverse of $k=x-\sin(x)$, together with a plot of the function itself (dashed line). Plotting them both on the same graph is useful, since they both have the same range for their independent and dependent variables. It can be observed that now there is a clear symmetry of this function at the point $(\pi,\pi)$, thanks to the removal of the square root term. This means that we only need to deal with the interval $[0,\pi]$, and use symmetry on that result to obtain the other half ($[\pi,2\pi]$).

Please see my accepted answer at Mathematics Stack Exchange, for some useful approximations of the inverse of $k=x-\sin(x)$, and of $\varphi\left(V\right)$. They may not be of much interest to pure mathematicians, but, being an electronics engineer, I consider them very useful in solving practical problems involving the inversion of $k=x-\sin(x)$ or similar functions.

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