Circular segment area inverse function - invert function of (x−sin(x)) Computing the area of a circular segment (green part below) from a distance along the radius axis (like $h$, the sagitta distance) is well-known.

What could be the inverse function, that is the $f$ function: $h=f(A)$, or an approximation of it ?
In other words: The circle is filled with a known quantity of a liquid ($A$) - what is the resulting liquid level ($h$) ?
EDIT 1
Considering that:
$h=R(1-\cos(\theta/2))$
and
$A=R^2/2.(\theta - \sin(\theta))$
I would simply need to express $\theta$ as a function of $A$, that is finding the invert function of $(x-\sin(x))$.
EDIT 2
According to this other question pointed out by Saad, it seems an algebraical expression of the inverse function cannot be obtained. An approximation would however be welcome.
EDIT 3
See the motivation behind this question here: https://observablehq.com/@jgaffuri/striped-circle
 A: A decent approximate for the sine function over its full variation domain $\theta\in[0,\pi]$ is 
$$\sin \theta = \frac{4\theta}\pi\left( 1- \frac \theta\pi\right)$$
(See the plot below.) Then, the area $A$ can be expressed as
$$A =\frac12 R^2\left( \theta - \sin\theta\right) = R^2\left[\left(\frac12-\frac2\pi\right)\theta + \frac2{\pi^2}\theta^2\right] $$
and the corresponding inverse function is
$$\theta(A) = \frac{\pi^2}4\left(\frac12-\frac2\pi\right) \left[\sqrt{1+\frac{32A}{(\pi-2)^2R^2}}-1\right]$$

A: Exact inversion is not possible because the equation is transcendental, so we have to resort to numerical methods.
We approximate the exact curve, in blue,
$$y=x-\sin x$$
in the range $[0,\pi]$. Other ranges follow by symmetry/periodicity.
Using the first two terms of the Taylor development, we get the green curve
$$y\approx \frac{x^3}6,$$ which matches the function well for small $x$.
For larger $x$, we can fix the deviation (up to $\dfrac{\pi^3}6$ instead of $\pi$) by adding a damping factor such that the function value is exact at $0$ and $\pi$. We also require that for small $x$, the formula remains asymptotically exact. For convenience, we chose the form $1-ax^3$ and obtained the magenta curve
$$y\approx\frac{x^3}6\left(1-\left(1-\dfrac6{\pi^2}\right)\frac{x^3}{\pi^3}\right).$$

As you can notice, both
$$y=ax^3$$ and $$y=ax^3(1-bx^3)$$ are easily inverted (for the second, solve the quadratic equation in $x^3$).
From these inital approximations, you can start Newton's iterations,
$$x_{n+1}=x_n-\frac{x_n-\sin x_n-y}{1-\cos x_n}.$$
A: Due to their transcendental nature an exact inverse relation cannot be defined. A numerical iteration supplies exact heights. 
In the following a differential equation approach is chosen and set up in  Mathematica. Started with parametrization on $t$ but ended up as normal ODE to output height $h$ as dependent variable. It is integrated to read off y-axis water heights at desired x-axis volume fraction of cylindrical tank ( three outputs shown) :
$$ x= \dfrac{A}{\pi R^2}=\dfrac{t-\sin t}{2 \pi};\, y= \dfrac{h}{R}=1-\cos t/2\, $$
$$ x'=\dfrac{1 - \cos t}{2 \pi}; \,y'= \dfrac12 \,\sin t/2 $$
$$ \dfrac{dy}{dx}=\dfrac{y'}{x'}=\pi \dfrac {\sqrt{1-(1-y)^2}}{2 ( 1- \cos^2 t/2})= \dfrac{\pi/2}{\sqrt{y(1-y)}}$$
 niveau = {Y'[x] == (Pi/2)/Sqrt[Y[x] (2 - Y[x])], Y[0] == 10^-8};
    NDSolve[niveau, Y, {x, 0, xm}];
    y[u_] = Y[u] /. First[%];
    Plot[y[x], {x, 0., 1}, GridLines -> Automatic, AspectRatio -> 1, 
     PlotStyle -> {Blue, Thick}, Axes -> True, 
     AxesLabel -> {AreaFraction, Height}]
    {y[0.2].y[0.6], y[0.85]}
    ParametricPlot[{(t - Sin[t])/(2 Pi), (1 - Cos[t/2])}, {t, 0, 2 Pi}, 
     AspectRatio -> 1, PlotStyle -> {Blue, Thick}, GridLines -> Automatic,
      Axes -> True, AxesLabel -> {AreaFraction, Height}]


Integrated and direct plots are seen to be identical.
