Symbolic Notation for $\theta "=" \arcsin(-.5)$? I'm teaching PreCalculus and the following issue has always bugged me.
Problem: Solve $\sin\theta = -.5$ for $0 \le \theta \le 2\pi$.
Solution:
\begin{align*}
\sin\theta &= -.5\\
\theta &= \arcsin(-.5) = -\frac\pi6
\end{align*}
But to get this into our desired domain, our solutions are $\boxed{\theta = \frac{7\pi}6 \text{ or }\frac{11\pi}6}$.
So my objection is the line $\theta = \arcsin(-.5)$, because that's really not true. $\theta$ can be a whole lot of things! So does there exist some symbol or notation that expresses this? Something a-la "If $x^2=9$, then $x = \pm 3$." Like
$$\theta \stackrel{\text{is related to}}{\sim} \arcsin(-.5) = -\frac\pi6 ?$$
 A: $x=\sin(\theta)$ is periodic (and therefore not injective) in $\theta$ so no inverse function can exist. But $\sin(\theta)$ domain limited to $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$ is injective (and surjective onto $-1\le x\le1$). The definition of $\theta=\arcsin(x)$ relies on the injectivity of $x=\sin(\theta)$ to be a proper inverse function. This restricts the range of $\arcsin(x)$ to $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$.
The equation $\sin(\theta)=-.5$ is has multiple solutions, so your students need to learn how to obtain the correct solution given a specific solution. A simpler example that you've demonstrate is $x^2=2$. What are the solutions to this equation? $+\sqrt 2$ and $-\sqrt 2$. But what is the range of $\sqrt{x}$? Just the positives. In order to find both solutions to the equation, you need to learn how to transform the singular solution from the square root function into the other.
In your problem, you went from $\sin(\theta)=-.5$ to $\theta=\arcsin(-.5)$. If it were $x^2=1$ going to $x=1$, you would have marked the student wrong, because the correct step is $x=\pm1$.
To concisely express all of the solutions to $\sin(\theta)=x$, you should write it out as $\theta=n\pi+(-1)^n\arcsin(x),\, n\in\mathbb Z$.
A: (Note: this would be a good question for Mathematics Educators SX.)
I absolutely agree with you: in the context of this question, the line "$\color{red}{\theta=\arcsin(-0.5)}$" is absolutely wrong and shouldn't be there.
From the point of view of providing a correct solution, I'm not aware of any special symbol for what you mean. I guess writing out the complete answer is an option, which can be done in the form
$$\theta=\arcsin(-0.5)+2\pi n=-\frac{\pi}{6}+2\pi n \; \text{and } \; \theta=\pi-\arcsin(-0.5)+2\pi n=\frac{7\pi}{6}+2\pi n,$$
or even in a more condensed form
$$\theta=(-1)^n\arcsin(-0.5)+\pi n=(-1)^n\frac{\pi}{6}+\pi n,$$
although the latter, while more concise, is less comprehensible. The next step would be to point out that of all those infinitely many solutions only two lie within $[0,2\pi]$, hence the answer.
From a pedagogical point of view, as a fellow teacher, I usually use a different approach for questions like this. I draw the unit circle, and ask my students to approach finding the solutions in two steps:


*

*First, identify in which two quadrants the solutions have to lie based on the fact that $\sin\theta$ is negative;

*Find the specific appropriate values of $\theta$ in those two quadrants to satisfy the equation $\sin\theta=-0.5$ specifically.


Using this approach I don't mention arcsine at all.
A: (The following is a didactical etude.)
We are told to give an explicit description of the set
$$S:=\bigl\{\theta\in{\mathbb R}\,\bigm|\,\sin\theta=-0.5, \ 0\leq\theta\leq2\pi\bigr\}=\sin^{-1}\bigl(\{-0.5\}\bigr)\cap[0,2\pi]$$in terms of the
$\arcsin$ function; the latter being available on our pocket calculator. By definition, for given $x\in[{-1},1]$ this function returns the unique angle $\alpha\in\left[-{\pi\over2},{\pi\over2}\right]$ satisfying $\sin\alpha=x$. In the case at hand $x=-0.5$, hence $\alpha=-{\pi\over6}$.
At this point the $\arcsin$ function has completely served its purpose, and the pocket calculator cannot further help us in solving this problem. Only our general knowledge about the $\sin$ function  can guide us further. This knowledge tells us that in any case $2\pi-{\pi\over6}\in S$, and as $\sin x\equiv\sin(\pi-x)$ we also have $\pi+{\pi\over6}\in S$. Further analysis, using the monotonicity of $\sin$ in the interval $\left[-{\pi\over2},{\pi\over2}\right]$, would then lead to the conclusion that there are no other solutions, so that in fact $S=\bigl\{{7\pi\over6},{11\pi\over6}\bigr\}$.
A: The way I always explained it to my students was "First, you find a solution, then you find the solution."
Let $\theta_0 = \arcsin(-\frac 12)$.

The range of $\arcsin$ is $-\frac{\pi}{2} \le \theta_0 \le \frac{\pi}{2}$. Since, on the unit circle, $\sin \theta = y=-\frac 12$, we see quickly that 
$\theta_0 = -\frac{\pi}{6}$. The two red dots indicate the two points on the unit circle for which $\sin \theta = -\frac 12$. One way to express this is
$\theta \in \{ -\frac{\pi}{6} + 2n\pi : n \in \mathbb Z\} \cup
            \{ \frac{7\pi}{6} + 2n\pi : n \in \mathbb Z\}$
To achieve $0 \le \theta \le 2\pi$ we find that
$\theta \in \{\frac{7\pi}{6}, \frac{11\pi}{6} \}$
where $ \frac{11\pi}{6} = -\frac{\pi}{6} + 2\pi$
