Some physical problem reduces to some equation that looks pretty transcendental:
$$ k \sin\mathopen{}\left(\sqrt{k^2+2\delta}\,\pi/2\right)\mathclose{}\cos(k\pi/2) + \sqrt{k^2+2\delta}\cos\mathopen{}\left(\sqrt{k^2+2\delta}\,\pi/2\right)\mathclose{}\sin(k\pi/2) =0. $$
with $\delta\ < k \in \Bbb R^+$.
For $\delta=0$ the solutions are $k\in\Bbb N$. What could one do for $\delta\ne 0$?
For $\delta \leq 0$, the equation LHS gives complex values when $|k| < \sqrt{-2\delta}$ and the only solution is $|k| = \sqrt{-2\delta}$.
The matter is much more interesting for $\delta > 0$. There is a set of discrete values of $\delta$ for which the equation can be solved exactly for $k$. Namely, if $\delta = 2n+2$ with $n \in \Bbb N$ then $k=n$ solves the equation, because then $\sqrt{k^2+2\delta} = k+2$ so the LHS becomes $$ k \sin \left(k \frac{\pi}2 + \pi \right) \cos \left(k \frac{\pi}2 \right) + (k+2) \cos \left(k \frac{\pi}2 + \pi \right) \sin \left(k \frac{\pi}2 \right) \\= (-k-(k+2)) \sin \left(k \frac{\pi}2 \right) \cos \left(k \frac{\pi}2 \right) = -(2k+2) \sin k\pi = 0 $$ For $\delta = 2n+2+\epsilon$, with $\epsilon$ small,we can do perturbation expansion, and the results are intriguing. When $n$ is even, it seems the function is zero at $$k = \frac{n-3}{2n+2} \epsilon + O(\epsilon^2)$$
But when $n$ is odd, it seems the solution value of $k$ moves away from its integer value much more rapidly as $\delta$ moves from its even integer value. The behavior is $$ k = \frac{n+2}{2} \sqrt{\epsilon} + \frac{n}{4n+4} \epsilon + O(\epsilon^{3/2})$$
and this covers the solutions on both sides of the except that there are no solutions for $0 < \delta < 4$.
This lack of solutions must have some physical meaning; I suspect that for these values of $\delta$ the energy is too small and the solution to the left of the boundary is a decaying exponential rather than a sine wave.