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?