Solving $(Ax + B)\sin x = C$ for $x$

Question

I have a pretty messy equation that I'm trying to solve for x. I've been able to get it down to:

$$(Ax + B)\sin x = C$$

Where $A,B,C$ are all constants. Is there an analytical solution to this?

Answer 1

As was pointed out on the remarks, if $A=0$ the equation reduces to the trivial $\sin x = C/B$, so let's assume $A \neq 0$.

One obvious simplification: WLOG, $C\in \{0,1\}$, otherwise, divide both sides by $C$ and rename $A/C \to A$ and $B/C \to B$.

Now if $C=0$ we get $(Ax+B)\sin x = 0$, which either implies $x = -B/A$ or $x \in \{n\pi\}_{n \in \mathbb{Z}}$.

The final case is indeed $(Ax+B)\sin x = 1$, which does not have easily expressible solutions. I would recommend either

• if you have a computer available, for the interesting set of $(A,B)$ compute it using Newton's Method (or Wolfram Alpha) to get numerical results.
• if no computer is available, you can eyeball it by drawing a graph of $\sin x$ and $\frac{1}{Ax+B} = \frac{1}{A(x+B/A)}$ (which is a scaled and translated hyperbola) on the same scale and then doing 1-2 iterations of Newton's Method from your eyeball point to get reasonable precision...