## Precalculus (6th Edition)

$\emptyset.$
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: $\emptyset.$