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

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?

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Assuming $A \neq 0$, there are infinitely many solutions for any $B$ and $C$. I do not think that these solutions will have a "nice" closed-form though. –  Lord Soth Jun 28 '13 at 19:16
@LordSoth: I see, thanks for the quick response. I think I may need to look back at how I got to this point then, given what I know about the application, there should only be one solution. –  GeneralMike Jun 28 '13 at 19:23
Even if $A=0$ there are infinitely many solutions (if $|\frac{C}{B}|\leq 1$), but the form is closed and nice :-) –  Avitus Jun 28 '13 at 19:32
If $A$ is nonzero you can rewrite in the form $\sin x =\frac c{x+b}$ and see the pattern of solutions by sketching $y=\sin x$ and $y=\frac c{x+b}$ on the same graph. If $A=0$ then the second graph is a horizontal line. Sometimes with equations in $\sin x$ there are other limits on the value of $x$ which restrict the number of solutions. –  Mark Bennet Jun 28 '13 at 19:32

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...