Answer
$\emptyset.$
Work Step by Step
As solved in exercise $5$,
the angle (degree measures) with sine equal to $-1$ is$ 270^{o}$.
Also, if $0 \leq \theta < 360^{o}\qquad/\div 2$
then
$0 \displaystyle \leq \frac{\theta}{2} < 180^{o}.$
But,
$\displaystyle \frac{\theta}{2}= 270^{o}$ is not in the interval $0 \displaystyle \leq \frac{\theta}{2} < 180^{o}$.
There are no solutions in the interval $0 \leq \theta < 360^{o}$.
Solution set: $\emptyset.$
