How was this approximation of transcendent equation solution found? I have an equation for $\xi$:
$$\xi\gamma=\cos\xi,$$
where $\gamma\gg1$. I've tried solving it assuming that $\xi\approx0$ and approximating $\cos$ by Taylor's second order formula:
$$\xi\gamma\approx1-\frac{\xi^2}2,\tag1$$
then I get
$$\xi\approx2\left(\sqrt{\gamma^2+1}-\gamma\right).\tag2$$
The book I read gives the approximate solution of it as
$$\xi\approx\frac1\gamma\left(1-\frac1{2\gamma^2}\right).\tag3$$
When I tried to understand how they got this, my first thought was like "they made another simplification and assumed $\xi\approx\frac1\gamma$ in $(1)$". I thus thought that this approximation would be worse. But when plotting these solutions as functions of $\gamma$, I found that $(3)$ in fact converges much faster to numerical solution than $(2)$!
So I now wonder: how did they get $(3)$?

$f(x)=\cos x-\gamma x$ is a concave decreasing function over $(0,\pi/2)$, hence Newton's method gives that the first positive root of $f(x)$ is less than:
$$ 0-\frac{f(0)}{f'(0)} = \frac{1}{\gamma} $$
as well as it is less than:
$$\frac{1}{\gamma}-\frac{f(1/\gamma)}{f'(1/\gamma)}=\frac{1}{\gamma}-\frac{1}{2\gamma^3}+O\left(\frac{1}{\gamma^5}\right).$$
Further iterations do not change the appearance of the last asymptotics.
