Approximate solution to a transcendental equation I'm working on a physics problem and stumbled upon the following equation:
$$h=\frac{2n\pi+\arctan\left(\frac{c}{b}\right)}{b}$$
where $n \in \mathbb{Z}$, $c \in [0,20]$ and $h \in \mathbb{R}^+$. This equation has to be solved for $b$, which can be done numerically ofcourse, but if possible, I would love to have a quite accurate expression in the given interval. Ive already did some work and obtained the following fair approximation for $b$:
$$b\approx\frac{(1+2n)\pi c}{2+ch}$$ However, this equation breaks down at low $c$. Thus, I was wondering if more accurate (analytical) expressions are possible.

We deal with the case $c > 0, h > 0, n\ge 1$.
The equation is written as
$$b = \frac{2n\pi}{h} + \frac{\arctan\frac{c}{b}}{h}.$$
Clearly, the equation has a unique real solution $b$. More precisely,
this solution is located in $ (\frac{2n\pi}{h}, \frac{2n\pi + \pi/2}{h})$.
Using the famous Lagrange inversion theorem, the solution is given by
$$b = \frac{2n\pi}{h} + \sum_{k=1}^\infty \frac{1}{k! h^k}
\frac{\partial^{k - 1}}{\partial x^{k - 1}}\left[\left(\arctan\frac{c}{x}\right)^k\right]\Bigg\vert_{x = \frac{2n\pi}{h} }.$$
An approximation is
\begin{align*}
 b &\approx \frac{2n\pi}{h} + \sum_{k=1}^2 \frac{1}{k! h^k}
 \frac{\partial^{k - 1}}{\partial x^{k - 1}}\left[\left(\arctan\frac{c}{x}\right)^k\right]\Bigg\vert_{x = \frac{2n\pi}{h} }\\
 &= \frac{2n\pi}{h} + \frac{1}{h}\arctan \frac{ch}{2n\pi}
 - \frac{c}{4\pi^2 n^2 + c^2h^2}\arctan \frac{ch}{2n\pi}.
\end{align*}
Examples:
(1) $c = 1/2, h = 1, n = 2$,
the solution is $\approx 12.60601344$,
and the approximation is $12.60601266$.
(2) $c = 1/100, h = 1/5, n = 10$,
the solution is $\approx 314.1594245$,
and the approximation is $314.1594246$.
(3) $c = 1/10, h = 100, n = 10$,
the solution is $\approx 0.6298929625$,
and the approximation is $0.6298929436$.
