Good way of explaining how to find all solutions to $\sin\theta+a=0$ I'm not sure exactly how to properly describe this trigonometry problem, so I will just write it out. It is less of a "problem" and more of a "I can't find a really good way to explain it".
We want to find a value $\theta$ such that $$\sin\theta + a = 0,$$ where $|a| < 1$. Subtracting $a$ from both sides gives us $\sin\theta = -a$, and one solution is easily seen to be $\theta = \arcsin(-a)$. However, because $-1 < -a < 1$, there should be two possible values of $\theta$ that give $\sin\theta = -a$. I'll call the one we already found $\theta_1$, and the one we are looking for $\theta_2$.
The best explanation I could think of is, when looking at the unit circle, $\sin\theta_1$ is considered the $y$ coordinate on the circle, while $\cos\theta_1$ is the corresponding $x$ coordinate. However, if you reflected across the $y$ axis, that other angle will also satisfy $\sin\theta_2 = -a$. This reflection across the $y$ axis corresponds to $\theta_2 = \pi - \theta_1$, and this is correct, because $$\sin\theta_2 = \sin(\pi - \theta_1) = \sin\pi + \sin\theta_1 = 0 -a = -a.$$ So we have our two solutions: $$\theta_1 = \arcsin(-a) \\ \theta_2 = \pi - \arcsin(-a).$$
My problem is I feel like my solution is too "wordy" and not elegant enough. Maybe it's not possible, but is there a better, more analytic way of deriving this result? Since $|\arctan x| \le \pi/2$, I feel like it would be impossible to easily derive $\theta_2$ like we did $\theta_1$, since $|\theta_2| \ge \pi/2$, so there is no way for $\arctan x$ to ever map to that value.

Let $y = \arcsin a $
So we have,
$$\sin x + \sin y = 2\sin\frac{x+y}{2} \cos\frac{x-y}{2} = 0 \Rightarrow \sin\frac{x+y}{2} 
 =0 \text{ or }\cos\frac{x-y}{2} = 0  $$
$$x+y = \text{even multiple of }\pi \text{ or } x - y = \text{odd multiple of }\pi$$
$$ \color{blue}x = \begin{cases}\text{even multiple of }\pi  - y \\\text{or} \\\text{odd multiple of }\pi + y\end{cases} = n\pi + (-1)^ny = \color{blue}{n\pi + (-1)^n\arcsin a} , n \in \Bbb Z$$
