#### Answer

$\{\ 270^{o} \ \}$

#### Work Step by Step

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: $\{\ 270^{o} \ \}$