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Here's the original equation:

$$\frac{1}{\beta}arctan\left(\frac{\sqrt{2rx+x^{2}}}{r}\right)+\left(r\sqrt{2rx+x^{2}}-r^{2}arctan\left(\frac{\sqrt{2rx+x^{2}}}{r}\right)\right)=\frac{\pi}{2\beta}$$

which I've expanded:

$$\frac{\sqrt{2}}{\beta\sqrt{r}}x^{1/2}+\left(\frac{8\beta r^{2}-5}{6\beta r^{3/2}\sqrt{2}}\right)x^{3/2}=\frac{\pi}{2\beta}$$

because the top expression only needs to be valid for small $x$.

Can you solve either expression for x? Perhaps by writing x = x0 + x1 and approximating a solution? I can't just get a huge mess, I'd much rather an expression that is slightly off but much simpler.

Cheers!

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    $\begingroup$ As Ross notes, this cubic has one nasty real root that can be found by wolframalpha, and one look at it should warn you that you shouldn't be expected to find it by hand. One thing that might yield a simplification (or at least something different) is whether you might mean "x is small compared to r" instead of merely just "x is small", as well as any information about $\beta$. $\endgroup$ – David H Oct 16 '13 at 17:45
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You have $ax^{1/2}+bx^{3/2}=c=x^{1/2}(a+bx)$ Then $c^2=x(a+bx)^2$ and you have a cubic which can be solved by the usual messy technique.

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