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($x \in \mathbb{R}$) 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.

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    Keep in mind "The positive solution to $\sin(2x) = x$" **is** an exact solution to your equation. It may even be useful for many applications.2012-12-08

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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} . $

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    I suppose given that there are few exact values of $\sin(2x)$, this is the best possible (I was asking too much for an exact solution to an equation including a function that doesn't have exact values for most $x$).2012-12-08
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Three solutions exist, as is apparent by the graphs of $x$ and $\sin (2x)$.

One solution is trivially $0$.

The other two solutions, $S_1$ and $S_2=-S_1$, which don't have elementary closed forms, can be obtained to sufficient accuracy with Newton's method (or another root-finding algorithm).

$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x+\frac{x-\sin (2x)}{2\cos(2x)-1}$

where the choice of $x_0$ will give one of the three different solutions.

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    Thanks- I'll leave this unaccepted for a while just in case a proof either way turns up.2012-12-08