Answer
$\{60^{o},135^{o},240^{o},315\}$
Work Step by Step
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= $\{60^{o},135^{o},240^{o},315\}$
