How to solve $\beta = \alpha - \arcsin(k \cdot \sin(\alpha))$ for $\alpha$?

Question

I'm looking for an analytical expression in the form of $\alpha = f (\beta)$, so solving the equation below for $\alpha$. Is this possible?

$$\beta = \alpha - \arcsin(k \cdot \sin(\alpha))$$

Answer 1

We have $$\arcsin(k\cdot\sin \alpha) = \alpha - \beta $$ so $$k\cdot \sin \alpha =\sin(\alpha - \beta) = \sin \alpha \cos\beta - \cos\alpha\sin\beta$$

Dividing by $\cos \alpha$ we get $$k\cdot \tan \alpha = \tan\alpha \cos\beta-\sin \beta$$

That is $$\tan\alpha = \frac{\sin\beta}{\cos\beta-k}$$ and finally $$\alpha =\arctan\left(\frac{\sin\beta}{\cos\beta-k}\right)$$

Since we are using inverse trigonometric functions this will be valid only on a certain interval and some adjustment will be necessary if one needs to extend the answer to a larger domain.