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I did this using trial and error, but I was just wandering if there is an algebraic way of solving this?

I thought about double angle formula but that doesn't work properly does it?

I then tried writing it in the form $x \cdot 2 \cot(x) = 1$ but even then, I can't solve it properly by re-writing $\cot(x)$ because of the $x$ outside the brackets.

Anybody know how I would do this algebraically?

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To begin with, a plot will show you that there is exactly one solution in each of the intervals $(-\pi/2 +k\pi,\pi/2+k\pi$. – 1015 Feb 13 '13 at 18:47
I seem to recall this question here already... – GEdgar Feb 13 '13 at 18:52
I've seen such an equation occur in the context of quantum mechanics, where the eigenvalues of some hamiltonian are given by the set of solutions to such an equation. I strongly suspect there is no "nice" way to find these solutions, however, insofar as there is no combination of elementary functions that'll give you the solutions (save $x = 0$). Instead, it's more interesting to develop asymptotics for these solutions. The idea is to approximate $\tan$ by vertical lines at $\frac{\pi}{2} + k \pi$ as $k$ ranges over the integers. – A Blumenthal Feb 13 '13 at 18:53
I work with such a functions pretty often, so I confirm - there's no "nice" solution for it. They can be found numerically though, and that's what we do. – Kaster Feb 13 '13 at 18:56
@Kaish: Here is a plot and here are numerical solutions to go along with the other comments. Regards – Amzoti Feb 13 '13 at 19:01
up vote -2 down vote accepted

It cannot be solved algebraically as it is a transcendental equation.

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There are trascendental equations that can be solved algebraically. For instance, $\sin x = \cos x$. – vonbrand Feb 14 '13 at 1:38

Of course, the roots are transcendantal numbers, which cannot be expressed in terms of a finite number of elementary functions. But they can be expressed on the forme of series :

enter image description here

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