#### Answer

$\{45^o,135^{o},225^{o},315^{o}\}$

#### Work Step by Step

Recognize the RHS of a double angle identity for cosine,
$\cos 2\theta=0.$
Since $0^{o} \leq \theta < 360^{o},$
then $0^{o} \leq 2\theta < 720^{o},$
So, $ 2\theta$ can be
$90^{o}, 270^{o}, $
$90^{o}+360^{o}=450^{o}$ or $270^{o}+360^{o}=630^{o}.$
Halving these angles,
$\theta=45,135^{o},225^{o},315^{o}$
Solution set = $\{45,135^{o},225^{o},315^{o}\}$
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Another method: recognize a difference of squares
$(\cos\theta+\sin\theta)(\cos\theta-\sin\theta)=0$
.. and apply the zero factor principle ...
1. $\cos\theta=-\sin\theta\quad $or
2. $\cos\theta=\sin\theta$,
yielding the same solution set.