# Is $\sin(2x)=x$ solvable?

($x \in \mathbb{R}$) 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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One of the solutions is exactly $0$. The other two are atractive fixed points of $\sin(2x)$, so just iterating the function gets you alternately upper and lower bounds on the solution, about $\pm0.94774713\ldots$. Newton's method would get it a lot faster. – Michael Hardy Dec 8 '12 at 20:04
What do you mean by "find the exact answer"? What sort of answer counts? – Chris Eagle Dec 8 '12 at 20:53
OK, the solutions are $0$, the unique positive real $a$ such that $\sin(2a)=a$, and $-a$. – Chris Eagle Dec 8 '12 at 21:00
This is a transcendental equation and as I know it can't be solved exactly, except for trivia solutions. – m0nhawk Dec 8 '12 at 21:13
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. – Hurkyl Dec 8 '12 at 22:30

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 $x$). – Alyosha Dec 8 '12 at 21:44

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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So is it impossible not to resort to computation? – Alyosha Dec 8 '12 at 21:09
@Alyosha I believe so, yes. Even both Mathematica and Sage cannot find closed forms. – Argon Dec 8 '12 at 21:11
Thanks- I'll leave this unaccepted for a while just in case a proof either way turns up. – Alyosha Dec 8 '12 at 21:12