Answer
As $-30^\circ\lt0^\circ$, $-30^\circ$ does not lie in the given interval $[0^\circ,360^\circ)$, so it cannot be the correct answer.
Work Step by Step
1) Summarize the exercise:
- The equation should be solved over the interval $[0^\circ,360^\circ)$
- We reach the step $\sin\theta=-\frac{1}{2}$
2) When it is stated that the equation is solved over the interval $[0^\circ,360^\circ)$, this means there is a condition for $\theta$, which is $$0^\circ\le\theta\lt360^\circ$$
As for $-30^\circ$, it is obvious that since $-30^\circ\lt0^\circ$, $-30^\circ\notin[0^\circ,360^\circ)$
Therefore, while $\sin(-30^\circ)$ does equal $-\frac{1}{2}$, it does not lie in the given interval, so it cannot also lie in the solution set, and the correct answers.
