## Precalculus (6th Edition)

$\{0^{o}, 45^{o},225^{o}\}$
Using the zero product principle, either 1. $\quad \tan\theta-1=0\quad$or 2. $\quad \cos\theta-1=0$ $1.\quad\tan\theta=1$ $\theta=45^{o}\quad$or $\theta=45^{o}+180^{o}=225^{o}$ $2.\quad\cos\theta=1$ $\theta=0^{o}$ Solution set = $\{0^{o}, 45^{o},225^{o}\}$