## Precalculus (6th Edition)

$\{60^{o},135^{o},240^{o},315\}$
Substituting t$=\tan\theta$ ,the equation becomes $t+1=\displaystyle \sqrt{3}+\frac{\sqrt{3}}{t}\qquad/\times t$ $t(t+1)=\sqrt{3}(t+1)$ $t(t+1)-\sqrt{3}(t+1)=0$ $(t+1)(t-\sqrt{3})=0$ ... apply the zero factor principle... $t=-1\quad$or $\quad t=\sqrt{3}$ Back-substitute: $1. \quad \tan\theta=-1\qquad$or $2. \quad \tan\theta=\sqrt{3}$ $1.\qquad\theta=135^{o}\quad$or $\theta=135^{o}+180^{o}=315^{o}$ $2. \qquad\theta=60^{o}\quad$or $\theta=60^{o}+180^{o}=240^{o}$ Solution set= $\{60^{o},135^{o},240^{o},315\}$