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?
It would be simpler to look for the zeros of function $$f(\omega)=\left(b^2 \omega ^2-1\right) \sin (\omega )-2 b \,\omega\, \cos (\omega )$$ which does not show any discontinuity.
Focusing on the first postive root, series expansion followed by power series reversion gives, as an approximation, $$\omega=t+\frac {20 b^2+10 b+1}{40(6 b^2+6 b+1 ) } t^3+$$ $$\frac {48720 b^4+45360 b^3+15300 b^2+2140 b+107}{67200(6 b^2+6 b+1 )^2}t^5+O(t^7)$$where $$t=\sqrt{\frac{6 (2 b+1)}{6 b^2+6 b+1} }$$
As shown below, the approximation is decent (it could be improved at the price of more terms in the initial series expansion). $$\left( \begin{array}{ccc} b & \text{estimate} & \text{solution} \\ 0.05 & 2.69706 & 2.85774 \\ 0.10 & 2.49284 & 2.62768 \\ 0.15 & 2.32792 & 2.43990 \\ 0.20 & 2.19140 & 2.28445 \\ 0.25 & 2.07603 & 2.15375 \\ 0.30 & 1.97688 & 2.04223 \\ 0.35 & 1.89049 & 1.94582 \\ 0.40 & 1.81433 & 1.86151 \\ 0.45 & 1.74654 & 1.78704 \\ 0.50 & 1.68568 & 1.72067 \\ 0.55 & 1.63066 & 1.66106 \\ 0.60 & 1.58059 & 1.60715 \\ 0.65 & 1.53479 & 1.55811 \\ 0.70 & 1.49267 & 1.51325 \\ 0.75 & 1.45378 & 1.47201 \\ 0.80 & 1.41772 & 1.43394 \\ 0.85 & 1.38417 & 1.39866 \\ 0.90 & 1.35285 & 1.36584 \\ 0.95 & 1.32353 & 1.33521 \\ 1.00 & 1.29600 & 1.30654 \\ \end{array} \right)$$
In any manner, with these estimates, Newton method converges in a couple of iterattions.