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I'm trying to find the roots $\omega_n$ of, \begin{equation} \tan(\omega) = \frac{2b\omega}{b^2\omega^2 -1} \end{equation}

I know it must be done numerically, so I would use the bisection method. In order to use that I must know the interval to search for each root for any $b \in (0,1)$. The problem is that there is different behaviour of the singularities at $b>1/2$. Does anyone know a way to solve the roots of this?

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    Have you plotted the functions? Perhaps you can gain insights as to their respective characteristics. Have you heard of Newton's method or the secant method? – 2017-02-11
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    Yes, I plotted and it can be seen here: https://postimg.org/image/f85kkxe8n/ The dashed lines are $b \in (1/2,1), solids are $b \in (0,1/2]$. Note where the singularities lie for them. – 2017-02-12
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    Which roots are you interested in? There are infinitely-many real roots. Could you also clarify what you mean by "there is different behavior of the singularities at $b > 1/2$"? I don't notice a difference between $b<1/2$ and $b>1/2$ in the plot you linked. – 2017-02-12
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    I would like to be able to find all up to an arbitrary $n$. The plot shows the first four roots. I'll clarify, @antonio-vargas. Except for the first three roots ($n=0,1,2$), the dashed lines ($b>1/2$) begin and end at $\pi/2 + n\pi$, sequential singularities of $\tan\omega$. For the solid ($b<1/2$) some lines begin and end at tan singularities and others asymptote to other intermediate points. Knowing the interval to search for a particular root is what I need for applying the bisection method. – 2017-02-13

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