Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - 7.6 Trigonometric Equations - 7.6 Exercises: 30

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= $\{60^{o},135^{o},240^{o},315\}$
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