## Precalculus (6th Edition)

$\{\ 270^{o} \ \}$
As solved in exercise 6, the angles (degree measures) with cosine equal to $-\displaystyle \frac{\sqrt{2}}{2}$ are $135^{o}$ or $225^{o}$ Also, if $0 \leq \theta < 360^{o}\qquad/\div 2$ then $0 \displaystyle \leq \frac{\theta}{2} < 180^{o}.$ $\displaystyle \frac{\theta}{2}=225^{o}$ can not be a solution as it does not belong to the interval $0 \displaystyle \leq \frac{\theta}{2} < 180^{o}.$ $\displaystyle \frac{\theta}{2}=135^{o}$ belongs to the appropriate interval, so $\theta=270^{o}$ is a solution Solution set: $\{\ 270^{o} \ \}$