Solving for $x$: $a \sin(bx + c) = \sin(x)$ Ok, I scoured the internet for more than a few months for this one (whenever I had the time). But just because of this question I've created my first Stack Exchange account.
I have thrown this at all Computer Algebra System software I was accessible to. (not Maple or Mathematica though)
I also tried complex transformation, no luck there (not too much experience there)
I know this could be easily numerically solved because it is guaranteed that there is exactly one solution for every interval when: $a \sin(bx + c) \le 1$. In my case, the general solution is crucial because I'm applying this to an optimization where I can't afford to have an additional dimension to be optimized.
This problem can be visualized as the intersection of two sine waves.
A simplified version of the problem (when a = 1) could be easily solved with the sum-to-product identity:
$\sin(bx + c) = \sin(x)$
$0 = \sin(bx + c) - \sin(x)$
$0 = 2 \cos(\frac{x(b + 1) + c}{2})\sin(\frac{x(b - 1) + c}{2})$
$0 = \cos(\frac{x(b + 1) + c}{2})$ and $0 = \sin(\frac{x(b - 1) + c}{2})$ etc...
But for the time being, the only valid assumption in my case is that b is rational
 A: COMMENT.-I doubt that there is a general closed solution as a function of $a, b, c$. However, I offer you here an equation of which approximate solutions could be found.
We have $$a\sin(bx+c)=a(\sin(c)\cos(bx)+\cos(c)\sin(bx))$$ from which we get the equation
$$a\sin(c)\sum_{k=0}^{\infty}\frac{(-1)^k(bx)^{2k}}{(2k)!}+a\cos(c)\sum_{k=0}^{\infty}\frac{(-1)^k(bx)^{2k+1}}{(2k+1)!}=\sum_{k=0}^{\infty}\frac{(-1)^k(x)^{2k+1}}{(2k+1)!}$$
This is a really difficult problem and what you want is probably impossible, I think.
A: This is a supplement to the answer of @Piquito. Recalling the identity theorem for power series, we make a coefficient comparison of the series expansion
\begin{align*}
a\sin(c)\sum_{k=0}^\infty (-1)^k\frac{\left(bx\right)^{2k}}{(2k)!}
+a\cos(c)\sum_{k=0}^\infty (-1)^k\frac{\left(bx\right)^{2k+1}}{(2k+1)!}
=\sum_{k=0}^\infty(-1)^k\frac{x^{2k+1}}{(2k+1)!}
\end{align*}
In the following we denote with $[x^n]$ the coefficient of $x^n$ of a series.

We consider even and odd coefficients separately and obtain for $n\geq 0$:
\begin{align*}
&[x^{2n}]:\quad &a\sin(c)\frac{(-1)^nb^{2n}}{(2n)!}&=0&\\
&&a\sin(c)b^{2n}&=0&\tag{1}\\
\\
&[x^{2n+1}]:\quad &a\cos(c)\frac{(-1)^nb^{2n+1}}{(2n+1)!}&=\frac{(-1 )^n}{(2n+1)!}&\\
&&a\cos(c)b^{2n+1}&=1&\tag{2}\\
\end{align*}

From (1) we conclude that at least one factor has to be equal to zero. Looking at (2) we see  that $a,b\ne 0$ and it follows
\begin{align*}
\sin(c)=0\qquad\rightarrow\qquad c=j\pi\qquad j\in\mathbb{Z}\tag{3}
\end{align*}
Putting $c=j\pi$ in (2) we obtain
\begin{align*}
a(-1)^jb^{2n+1}=1\qquad\qquad n\geq 0,\ j\in\mathbb{Z}\tag{4}
\end{align*}
From (4) we see that $j\in2\mathbb{Z}$. Since (4) has to be valid for all $n\geq 0$ we conclude that $b=1$ and finally that $a=1$.

Conclusion: The equation
\begin{align*}
a\sin(bx+c)=\sin(x)\qquad\qquad a,b,c\in\mathbb{C}
\end{align*}
has a solution if and only if $a=b=1$ and $b\in 2j\pi, j\in\mathbb{Z}$. In this case we have the identity
\begin{align*}
\sin(2j\pi x)=\sin(x)\qquad\qquad x\in\mathbb{C}
\end{align*}

