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 - Page 721: 31

Answer

$\{45^{o},135^{o},225^{o},315^{o}\}$

Work Step by Step

Substituting t$=\tan\theta$ ,the equation becomes $t-\displaystyle \frac{1}{t}=0,\qquad t\neq 0$ $t-\displaystyle \frac{1}{t}=0\quad/\times t$ $t^{2}-1=0$ ... difference of squares ... $(t+1)(t-1)=0$ ... apply the zero factor principle... $ t=1\quad$or$ \quad t=-1$ $\tan\theta=1\quad$or$ \quad\tan\theta=-1$ $\theta=45^{o},225^{o}\quad$or$ \quad \theta=135^{o},315^{o}$ Solution set= $\{45^{o},135^{o},225^{o},315^{o}\}$
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