I'm looking for the inverse of the function (family) $$ f(\phi)=\phi + \frac{A}{B}\cdot\arcsin\left(B\cdot\sin(\phi)\right),$$ where $A=\sin\theta,$ and $B=\cos\theta$, where $\theta$ can be thought of as a parameter (i.e. constant).
My hunch is that this has no solution that can be written down using elementary functions. Now, $f$ is well-behaved, continuous and monotonically increasing, so there should be no problem making an algorithm to find the inverse.
Is this a function that is known to not have an inverse that can be written down using elementary functions?
What is the "best" way to invert this function using an algorithm?
This is not a complete answer at all.
I should not expect a closed form for the inverse.
I think that is would be better to not consider $A,B$ as parameters but to use their definition. So $$f(\phi)=\phi+\tan (\theta ) \sin ^{-1}(\cos (\theta ) \sin (\phi ))$$ If you are concerned by small values of $x$, you could make a Taylor expansion and then use series reversion. Around $x=0$, this would give $$\phi=\frac{1}{\sin (\theta )+1}f(\phi)+\frac{ \sin ^3(\theta )}{6 (\sin (\theta )+1)^4}f(\phi)^3+O\left(f(\phi)^5\right)$$
I wonder if it could or not be better to consider instead $$g(\phi)=f(\phi)\cos (\theta )=\phi \cos (\theta )+\sin (\theta ) \sin ^{-1}(\cos (\theta ) \sin (\phi ))$$ for which the same process would give $$\phi=\frac{\sec (\theta )+(\sin (\theta ) (\sin (\theta )+3)+3) \tan (\theta )}{(\sin (\theta )+1)^4}g(\phi)+\frac{ \tan ^3(\theta )}{6 (\sin (\theta )+1)^4}g^3(\phi)+O\left(g(\phi)^5\right)$$
For sure, the same could be done building the series around $\phi=a$ but the formulae are really messy to be reported here.
May I suggest you play with that ? And let me know.
If it works more or less, we could improve using instead $[1,n]$ Padé approximant.