Let's start by rewriting the inequalities:
$$\pi - \theta_\delta < \Bigl( \bigl[ k(\pi - 2\theta_\mathrm{i}) \bigr] \ \mathrm{mod} \ 2\pi \Bigr) < \pi + \theta_\delta$$
to eliminate redundant parameters, bearing in mind that $\theta_\delta,\theta_\mathrm{i}$ are in $[0,\pi/2]$.
The basic idea is to divide through by $2\pi$. To justify this we need a
Lemma For real numbers $x\ge 0$:
$$ ( x \bmod 2\pi ) = 2\pi \left( \frac{x}{2\pi} \bmod 1\right) $$
This is a good point to explain how the "mod" operation with real numbers works differently from how the integer modulo notation works. Taking the integers modulo a (fixed, nonzero) integer gives us a (finite) ring of integer residues by way of a congruence relation that preserves addition and multiplication.
It is a different case when real numbers are involved, which is discussed in the Answers to this Question: If Integers modulo n exist, do Reals modulo m exist? In brief only the additive structure of the real numbers is preserved by the corresponding equivalence relation.
Note also that the above Lemma is framed in terms of an operator $\bmod$ rather than in terms of an equivalence relation. So to be clear we will begin with defining the operator on positive real numbers and proving the required Lemma.
Definition For any real numbers $x\ge 0, y\gt 0$, let $x \bmod y$ be the value in $[0,y)$ given by $y$ times the fractional part of $x/y$, i.e. $$ \{x/y\}\equiv (x/y) - \lfloor x/y \rfloor$$
We limit our definition to nonnegative values $x$ because that meets our immediate purpose and avoids sorting through the conflicting ways other authors extend the operation to negative numbers. (See the linked Wikipedia article in our definition for more details.) We ask the Reader to verify that the calculation defined above produces a unique value in $[0,y)$.
Observe $( \frac{x}{y} \bmod 1 )$ produces a value in $[0,1)$, namely $1$ times the fractional part of $\frac{x}{y}$. Multiplying by $y$ then gives the precise value in $[0,y)$ required by our definition. The Lemma is then an immediate application of the definition and this observation.
Using the Lemma with $x = k(\pi - 2\theta_\mathrm{i})$ and $y = 2\pi$ allows us to divide the original inequalities by $2\pi$:
$$ \frac{\pi - \theta_\delta}{2\pi} \lt \left(k\frac{\pi - 2\theta_\mathrm{i}}{2\pi} \bmod 1 \right) \lt \frac{\pi + \theta_\delta}{2\pi} $$
Substituting $0.5 - \epsilon = \frac{\pi - \theta_\delta}{2\pi}$, $0.5 + \epsilon = \frac{\pi + \theta_\delta}{2\pi}$, and $\alpha = \frac{\pi - 2\theta_\mathrm{i}}{2\pi}$ results in the simplified parameters we sought:
$$ 0.5 - \epsilon \lt (k\alpha \bmod 1) \lt 0.5 + \epsilon \tag{*}$$
Note that integer $k$ has an essentially unchanged role in satisfying this pair of inequalities. By construction: $1/4 \le 0.5 - \epsilon \le 0.5 + \epsilon \le 3/4$, with equality in the middle if and only if $\epsilon = \theta_\delta/(2\pi) = 0$ (in which case the condition $(*)$ cannot be satisfied because it requires strict inequality). So henceforth let's assume $\epsilon \gt 0$, equiv. $\theta_\delta \gt 0$.
As Paul Sinclair has commented, the satisfaction of $(*)$ is always possible when $\alpha$ is irrational, and may or may not be possible if $\alpha$ is rational. There is a classic result in Diophantine approximation that with irrational $\alpha$, the limiting fraction of integers $k$ satisfying $(*)$ is $2\epsilon$ (see Corollary 6 of this nice survey paper by Li Du and its references).
If $\alpha = 0$, corresponding to $\theta_\mathrm{i} = \pi/2$, then no solution $k$ is possible. Otherwise whether rational $\alpha$ will give rise some solutions $k$ can be worked out by elementary arithmetic. For example, if $\theta_\mathrm{i} = 0$, then $\alpha = 1/2$ and given $\epsilon > 0$ the least solution would be $k=1$. On the other hand if $\alpha = 1/3$, then it is possible that $\epsilon$ may be close enough to zero to exclude all of $\alpha$'s integer multiples $k/3$ (e.g none between $0.4$ and $0.6$).
