$\def\k{\operatorname k}\def\W{\operatorname W}$
Using a Dirac $\delta(x)$ Fourier series and this post
$$\begin{align} y\cot(y)=\frac1x\implies\frac1{\sec^2(y)-x}=\int_0^\frac\pi2\delta(\tan(t)-xt)dt-\frac1{2|x-1|}=\frac14+\frac1\pi\sum_{n=1}^\infty\int_0^\frac\pi2\cos(n (\tan(t)-xt))dt-\frac1{2|x-1|}\end{align}$$
Now remember the Bateman k$_v(z)$ and Whittaker W$_{a,b}(x)$ functions:
$$\k_v(x)=\frac{\W_{\frac v2,\frac12}(2x)}{\left(\frac v2\right)!}=\frac2\pi\int_0^\frac\pi2\cos(x\tan(t)-vt)dt$$
Therefore, we hopefully solve for the smallest positive solution:
$$\frac{\tan(y)}y=x\implies y=\sec^{-1}\left(\sqrt{\frac1{\frac14+\frac12\sum\limits_{n=1}^\infty\frac{\W_{\frac{nx}2,\frac12}(2n)}{\left(\frac{nx}2\right)!}-\frac1{2|x-1|}}+x}\right)= \sec^{-1}\left(\sqrt{\frac1{\frac14+\frac12\sum\limits_{n=1}^\infty\k_{nx}(n)-\frac1{2|x-1|}}+x}\right) \tag1$$
There appears to be no other way to use the Bateman function to solve $\frac{\tan(y)}y=x$. A numerical tester with $y\to\sec^{-1}\left(\sqrt{\frac1y+x}\right)$ is inconclusive. Even though the above series is slow, it still could be a solution.
Is $(1)$ true?