Asymptotic behavior of the solutions to $\sin x = (\log x)^{-1}$ I apologize in advance for the length.
The equation $\sin x = (\log x)^{-1}$ has exactly one solution $x_n$ in the interval $(2\pi n,2\pi n + \pi/2)$ for $n \geq 1$, and the exercise (de Bruijn, Asymptotic Methods in Analysis, ch. 2) asks me to show that
$$
x_n = 2\pi n + (\log 2\pi n)^{-1} + O((\log 2 \pi n)^{-3}).
$$
To start, we have $0 < x_n - 2 \pi n < 1$ and
$$
\sin x_n = \sin (x_n - 2 \pi n) = (\log x_n)^{-1} \to 0,
$$
so $x_n \to 2 \pi n$.  It would make sense then to make the substitution $x_n = z + t$, where $t = 2 \pi n$, so that we're now concerned with finding the asymptotic behavior of $z$ in terms of $t$ as $t \to \infty$ in the equation
$$
\sin z = (\log (z + t))^{-1}.
$$
That is, we want to show that
$$
z = (\log t)^{-1} + O((\log t)^{-3}).
$$
So far I've only been able to show that $z = O((\log t)^{-1})$ through arguments which are probably not sound.
I've tried to apply the Lagrange Inversion Formula but I can't seem to get it into the right form.  If we let $w = \log t$, then $w = \frac{z}{f(z)}$, where
$$
f(z) = \frac{z}{\log(e^{1/\sin z} - z)}.
$$
But $f(0) = 0$ (and, probably more importantly, $f$ isn't analytic at $0$), so I can't apply Lagrange.  Of course there may be a "correct" way to rearrange the equation to put it into Lagrange form.
I've also considered applying Newton's method, but I don't know if that's valid.  Applying the method to $(\log (z + t))^{-1} - \sin z$ with $z_0 = 0$ I get
$$
z_1 = - (\log t)^{-1} ( 1 + O((t \log t)^{-2})),
$$
which at least has the right asymptotic behavior in the first term.  Trying to iterate using, for example, $x_0 = (\log t)^{-1}$ in the hopes of getting more stable terms leads me to a wall of computation, and I doubt that's the goal of the problem.  More importantly, even if I did get a stable asymptotic series as I continued to iterate, I don't know whether I'm actually converging to the actual root of the equation.
Lastly I should mention that I've also tried letting $z = x_n(2 \pi n)^{-1} - 1$, but this didn't seem to lead to anywhere helpful.
Any tips?
 A: So... $\sin(z_n)=(\log(2\pi n+z_n))^{-1}$ and you proved that $z_n=o(1)$. 
This implies that $\log(2\pi n+z_n)=\log(2\pi n)+\log(1+z_n/(2\pi n))=\log(2\pi n)+o(1/n)$ and that $\sin(z_n)=z_n+O(z_n^3)$. Hence, using the notation $\ell_n=(\log(2\pi n))^{-1}$,
$$
z_n+O(z_n^3)=\frac1{1/\ell_n+o(1/n)}=\frac{\ell_n}{1+o(\ell_n/n)}=\ell_n+o(\ell_n^2/n).
$$ 
Keeping only the terms of higher order, this proves that $z_n=\ell_n+o(\ell_n)$. Keeping all the terms, this proves that
$$
z_n=\ell_n+O(\ell_n^3)+o(\ell_n^2/n).
$$
Finally, since $1/n\ll\ell_n$, the term $o(\ell_n^2/n)$ is also $O(\ell_n^3)$ and you are done.
A: My idea is to use Newton's formula $x_{k+1} = x_k - \frac{f(x_k)}{f'(x_{k})}$. We start with $x_0 = z_n = 2 \pi n$ and $f(x) = \log(x) \sin(x)-1$. We get $x_1 = z_n + \ell_n^{-1}$ with $\ell_n = \log(2 \pi n)$. Next 
$$
  x_2 = z_n + \ell_n^{-1} -\frac{\left(z_n \ell _n+1\right) \left(\sin \left(\frac{1}{\ell _n}\right) \log \left(z_n+\frac{1}{\ell _n}\right)-1\right)}{\left(z_n \ell
   _n+1\right) \cos \left(\frac{1}{\ell _n}\right) \log \left(z_n+\frac{1}{\ell _n}\right)+\ell _n \sin \left(\frac{1}{\ell _n}\right)} 
$$
The correction term is of order $(6 \ell_n^3)^{-1}$.
