Solve transcendental equation: At $α\ll 1$, $\left|\cos x + \alpha \frac{\sin x}{x}\right| > 1$, determine the width of of the $k$-th zone at $k\gg1$. To solve this transcendental equations approximately:
At $\alpha \ll 1$ find the positive solution of inequality:
$\left|\cos x + \alpha \frac{\sin x}{x}\right| > 1$, they are divided into series of zones numbered with the integres: $k = 0, 1, \ldots$. Determine the width of of the $k$-th zone at $k \gg 1$.
My approach: if we multiphy both sides by $x$ will we get any advantage?
I need some hints for this problem: How to procceed next?
 A: By the usual trigonometric identities
$$
\cos(x)+\sin(x)\fracαx=\sqrt{1+\frac{α^2}{x^2}}\cos(x+\phi(x)),~~\phi(x)\approx -\fracαx
$$
If $x\approx k\pi$, $|k|>1$, then the width of the region around that point is approximately the distance of the roots of
$$
\cos(u)=\sqrt{1+\frac{α^2}{(k\pi)^2}}^{-1}\approx 1-\frac12\frac{α^2}{(k\pi)^2}+\frac38\frac{α^4}{(k\pi)^4}
$$
for the roots closest to $u=0$. In first order, this will be $u=\pm\fracα{k\pi}$, as $\cos(u)=1-\frac12u^2+\frac16u^4+...$, so that the width is $\frac{2α}{k\pi}$.
For $k=0,\pm1$, perhaps only for $k=0$, one would have to use another approach.

condensed approach One could condense the several approximations by using $\frac{α}{x}=\tan\phi(x)\approx\tan(\frac{α}{x})$ from the start. Then the inequality reduces to
$$
\left|\cos\left(x-\phi\right)\right|< \cos\left(\phi\right)
\iff
|x-k\pi|<\frac\pi2~\land~\cos\left(x-k\pi-\phi\right)< \cos\left(\phi\right)
$$
This now can be solved easily
$$
0<x-k\pi<2\phi(x)=2\arctan\left(\frac{α}{x}\right)
$$
where the upper bound can be approximated as above.
