Hi Eric!

You can rewrite tan x = A + B sin x as:$\displaystyle {\sin x \over \pm \sqrt{1-\sin^2 x}} = A + B \sin x$

Substitute y=sin x, and you get:$\displaystyle {y \over \pm \sqrt{1-y^2}} = A + B y$

which yields:$\displaystyle y^2 = (A + B y)^2 (1 - y^2)$

This is a 4th order polynomial, which can be solved as explained for instance here: Quartic function - Wikipedia, the free encyclopedia

This may be more involved than you'd like, although I have to say that the article makes it appear more difficult than it actually is.

Afterward, you get the answer with $\displaystyle x = \arcsin y$.

Is this what you had in mind, or were you hoping for an easier solution?