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 $(3)$?
 A: $f(x)=\cos x-\gamma x$ 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.
A: Since you already received the good answers in comments and answers, let me show you another approximation based on the fact that, for $-\frac{\pi}{2}\leq \xi \leq \frac{\pi}{2}$, $$\cos(\xi) \simeq \frac{5 \pi ^2}{\xi ^2+\pi ^2}-4$$ So, consider the solution of $$ \xi \gamma=\frac{5 \pi ^2}{\xi ^2+\pi ^2}-4$$ that is to say $$\gamma  \xi ^3+\pi ^2 \gamma  \xi -4 \xi ^2+\pi ^2=0$$ Using Cardano,it seesm that there is only one real root as soon as $$\gamma \gt \frac{\sqrt{\frac{13 \sqrt{65}}{8}-\frac{83}{8}}}{\pi }$$ this root $$\xi= \frac{A \left(2^{2/3} A+8\right)+2 \sqrt[3]{2} \left(16-3 \pi ^2 \gamma ^2\right)}{6
   A \gamma }$$ $$A=\sqrt[3]{128-63 \pi ^2 \gamma ^2+3 \sqrt{3} \pi  \sqrt{4 \pi ^4 \gamma ^6+83 \pi ^2
   \gamma ^4-256 \gamma ^2}}$$ can now be expanded for large values of $\gamma$ and the solution is approximated by $$\xi=\frac{1}{\gamma }-\frac{5}{\pi ^2 \gamma ^3}+\frac{55}{\pi ^4 \gamma
   ^5}+O\left(\left(\frac{1}{\gamma }\right)^6\right)$$ Limiting to the first terms you then have $$\xi\simeq\frac{1}{\gamma }-\frac{5}{\pi ^2 \gamma ^3}=\frac{1}{\gamma }\Big(1-\frac{5}{\pi ^2 \gamma ^2}\Big)$$ and see how close is $\frac{5}{\pi ^2}$ to  $\frac{1}{2}$.
Let me consider an example with a small value $\gamma=2$; the approximation of the book is then $0.437500$, the one I proposed is $0.436674$, the solution of the cubic equation $0.449785$ and the exact solution is $0.449785$.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{}$
\begin{align}
\xi = {\cos\pars{\xi} \over \gamma}
\end{align}

It leads to the iteration:
  $$
\xi_{n + 1} ={\cos\pars{\xi_{n}} \over \gamma}\,,\qquad \xi_{0} = 0
$$

$$
\xi_{1} = {1 \over \gamma}\,,\qquad\xi_{2}={1 \over \gamma}\,\cos\pars{1 \over \gamma}\approx\color{#66f}{\Large{1 \over \gamma}\pars{1 - {1 \over 2\gamma^{2}}}}
$$
since $\ds{\cos\pars{x} \approx 1 - \half\,x^{2}}$ when $\ds{x \approx 0}$.
