TITLE: Is $\sin(2x)=x$ solvable?
QUESTION [5 upvotes]: ($x \in \mathbb{R}$) Graphically, it's obvious that the equation should have 3 solutions for x, but I can't think of any way to solve this without resorting to computation of [the Maclaurin series for $\sin(2x)$]$\div x$ or some cleverer computational trick.
I considered representing $\sin(2x)$ as 
$\left(1-\frac{x}{\frac{1}{2} \pi}\right)\left(1+\frac{x}{\frac{1}{2} \pi}\right)\left(1-\frac{x}{\frac{2}{2} \pi}\right)\left(1+\frac{x}{\frac{2}{2} \pi}\right)...$, but that seems too daunting to be of any use.
Edit: Note that I'm looking for a way to find the exact answer (i.e. not just an approximation), or a proof that it's impossible to find.

REPLY [3 votes]: How about
$$
x = \frac{\pi}{2} \mathrm{sinc}^{-1}\left(\frac{1}{2}\right)
$$
where $\mathrm{sinc}$ is defined by
$$
\mathrm{sinc}(u) = \frac{\sin(\pi u)}{\pi u} .
$$