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I am struggling to remember what I need to do/can do to reduce this further, and searching hasn't helped much, probably since I am not sure how to word the problem correctly. Anyway, I am attempting to re-arrange this function to solve for x:

$$ n=\frac{sin(2x)}{2}-x $$

I'm thinking that I am just forgetting some simple thing, but it's been a while since I had to do any real math.

bss36504
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  • Are you solving for $x$ in terms of $n$? – m1cky22 Aug 04 '16 at 19:22
  • @m1cky22 yes. n is composed of other known constants, so I simplified it for this question. – bss36504 Aug 04 '16 at 19:23
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    This might be relevant http://mathcentral.uregina.ca/qq/database/QQ.09.00/roble1.html – m1cky22 Aug 04 '16 at 19:27
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    @m1cky22 Thanks a ton! I was beginning to wonder if it was iteratively calculated. Wolfram could give me numerical results, but no solution. I can write a program to do what I need with "good enough" accuracy. – bss36504 Aug 04 '16 at 19:30

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You can't solve it explicitly in terms of elementary functions. If you know $x$ is small, you can expand $\sin (2x) \approx 2x - \frac 1{3!}(2x)^3 + \frac 1{5!}(2x)^5 + \ldots$. The first term cancels with the $-x$ you have, so you get $n \approx \frac 86x^3, x \approx \sqrt[3]{\frac 34 n}$. If $n$ is rather large, you can use that $-\frac 12 \le \frac {\sin (2x)}2 \le \frac 12$ so $x \approx -n$ and do some root finding from there.

Ross Millikan
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  • Unfortunately, x will be anywhere from 0 to Pi. Interesting solution nonetheless, thank you. – bss36504 Aug 04 '16 at 19:33
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    Then you are into one dimensional root finding, which is what Alpha is doing. There are chapters in all numerical analysis texts. I like Numerical Recipes chapter 9, which is free on line. – Ross Millikan Aug 04 '16 at 19:57
  • Source: The power series of sin(x) is defined when asking the following question. http://math.stackexchange.com/questions/347648/infinite-factorization-power-series-of-sinx To prevent confusion for those on an algebra, precalculus level it's probably easier to accept the need for numerical, nonexact answers. – nickalh Aug 04 '16 at 21:18