If $\epsilon \in (0,1]$ and $\tan^2\theta \leq \epsilon$, then what are the admissible values of $\theta$ in terms of $\epsilon$?

Question

Let $\epsilon \in (0,1]$, and $\theta$ is such that $\tan^2\theta \leq \epsilon$. What are the admissible values of $\theta$ in terms of $\epsilon$?

The motivation for the question is the following:

If a point $p$ makes an angle $\theta$ with x-axis, so that the above inequality holds, then it must be in a double cone with x-axis as its axis. I am trying to figure out its angular radius.

Answer 1

$$\tan^2\theta \leq \epsilon \iff -\sqrt {\epsilon}\le \tan \theta \leq \sqrt {\epsilon}$$

You have infinitely many solutions with the central one $$-\tan ^{-1}\sqrt {\epsilon} \le \theta \le \tan ^{-1}\sqrt {\epsilon} $$

Other solutions are the shifts of this interval by multiples of $\pi$ in both directions.