## Precalculus (6th Edition)

$\{45^o,135^{o},225^{o},315^{o}\}$
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 = $\{45,135^{o},225^{o},315^{o}\}$ ------- 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.