Numerical inverse of a function How to approximate the inverse of the function below?
$$f(x) = \frac34 x - \frac 12\sin(2x) + \frac 1{16} \sin(4x)$$
The goal is to get $x$ values (range $[0, \pi]$) from values of $f$. The function doesn't seem to have analytical inverse, so I figured to approximate $f$ with a Taylor series, truncate, then try taking an inverse, eg with Mathematica's InverseSeries?
http://reference.wolfram.com/language/ref/InverseSeries.html
Does anyone have a different approach, perhaps without Mathematica? 
 A: You could try a root-finding method. If you have that $c$ is an $f$-value, then you can solve $h(x) = f(x) -c =0$ on $[0,\pi]$ via Newton's method or the secant method, for example.
A: Ironically, you can get a closed, quantile special function, inverse using this special case of Incomplete Beta function with Mathematica’s Inverse Beta Regularized, but parameters beyond $\frac52$ produce very specific equations:
$$\text B_{\sin^2(x)}\left(\frac52,\frac12\right)=\frac{3x}4-\frac12\sin(2x)+\frac1{16}\sin(4x)\implies \boxed{x\mathop=^{0\le x\le\frac{3\pi}8}_\text{one period}\sin^{-1}\sqrt{\text I^{-1}_\frac{8x}{3\pi}\left(\frac52,\frac12\right)}}$$
Use the periodicity of the original function to extend the domain of the inverse function:

Proof of result. Please correct me and give me feedback!
The error of the Taylor series to $n$ terms is given by:
$$\text B_{\sin^2(x)}\left(n+\frac52,\frac12\right)$$
Therefore, the inverse of the error of the Taylor series of $$\frac{3x}4-\frac12\sin(2x)+\frac1{16}\sin(4x)$$ to $n$ terms experimentally on $0\le x\le\frac{3\pi}8$ is:
$$\sin^{-1}\sqrt{\text I^{-1}_\frac{8x}{3\pi}\left(n+\frac52,\frac12\right)} $$
