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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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    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.2012-12-08
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    What do you mean by "find the exact answer"? What sort of answer counts?2012-12-08
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    Exact in the way that $\sqrt{2}$ is, and $1.414$ is not; that is, an answer that, as written (in finite time) on paper, we know to infinite accuracy.2012-12-08
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    OK, the solutions are $0$, the unique positive real $a$ such that $\sin(2a)=a$, and $-a$.2012-12-08
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    I didn't specify that $sin(2x)=x$ and $sin(2x)=-x$, I was more looking for the 'exact' form of the number $\pm 0.94774713...$.2012-12-08
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    This is a transcendental equation and as I know it can't be solved exactly, except for trivia solutions.2012-12-08
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    @monhawk: Ok, what are *you* using "solved exactly" to mean?2012-12-08
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    @ChrisEagle: Under "solved exactly" I mean finding such $x$ that is equals to **finite** number of any combination of elementary functions.2012-12-08
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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

2 Answers 2

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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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    So is it impossible *not* to resort to computation?2012-12-08
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    @Alyosha I believe so, yes. Even both Mathematica and Sage cannot find closed forms.2012-12-08
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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