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I only see numerical approaches to solve this equation. Is there an analytical solution to solve $x$ as a function of $y$ for the range $(0,2 \pi)$? If there is no solution, is it possible to proof it?

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What do you mean by 'solve'? Are you trying to find zeroes? –  Joe Johnson 126 Jun 7 '12 at 18:15
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I would be extremely surprised if there were. –  Harald Hanche-Olsen Jun 7 '12 at 18:15
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@JoeJohnson126: My guess is that he wants to solve for $x$ as a function of $y$. –  Harald Hanche-Olsen Jun 7 '12 at 18:16
    
@ Harald Hanche-Olsen: thats right –  varantir Jun 7 '12 at 18:24
    
Probably no solution in elementary functions. See, for example, Kepler's equation which is known to have no solution in elementary functions: en.wikipedia.org/wiki/Kepler's_equation –  GEdgar Jun 7 '12 at 18:39

1 Answer 1

For a full and complete answer you might want to look into inversion of the power series, although I have not checked the inverse function theorem conditions for this one thoroughly. Nevertheless here's my $O(y^5)$-worth take on it: $$y=\frac{\frac{x^{3}}{6}+O\left(x^{5}\right)}{\frac{x^{2}}{2}+O\left(x^{4}\right)}=\frac{2}{x^{2}}\frac{\frac{x^{3}}{6}+O\left(x^{5}\right)}{1+O\left(x^{2}\right)}=\frac{2}{x^{2}}\left(\frac{x^{3}}{6}+O\left(x^{5}\right)\right)\left(1+O\left(x^{2}\right)\right)=\frac{2}{x^{2}}\left(\frac{x^{3}}{6}+O\left(x^{5}\right)\right)=\frac{x}{3}+O\left(x^{3}\right)$$ $$x=3y+O\left(y^{3}\right)$$ $$x=\left(1-\cos x\right)y+\sin x=\frac{3y^{3}}{2}+O\left(y^{5}\right)+3y+O\left(y^{3}\right)-\frac{1}{6}\left(27y^{3}+O\left(y^{5}\right)\right)=3y-3y^{3}+O\left(y^{5}\right)$$

I am not entirely (neither meromorphically) confident about the last term at $\frac{1}{6}$ in brackets in the last line, however Grapher tells me I am not too far from truth enter image description here

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