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: 41

Answer

$\{106.3^{o}, 149.6^{o},286.3^{o},329.6^{o}\}$

Work Step by Step

Substitute t=$\tan\theta$ $t^{2}+4t+2=0$... quadratic formula... $t=\displaystyle \frac{-4\pm\sqrt{16-8}}{2}=\frac{-4\pm 2\sqrt{2}}{2}$ $=\displaystyle \frac{2(-2\pm\sqrt{2})}{2}=-2\pm\sqrt{2}$ $\tan\theta=-2\pm\sqrt{2}$ (tan is negative in quadrants II and IV) Calculator: $\tan^{-1}(-2+\sqrt{2})\approx$106.324949937$\approx 106.3^{o}$ and, $106.3^{o}+180^{o}=286.3^{o}$ Calculator: $\tan^{-1}(-2-\sqrt{2})\approx$149.6388065957$\approx 149.6^{o}$ and, $149.6^{o}+180^{o}=329.6^{o}$ Solution set = $\{106.3^{o}, 149.6^{o},286.3^{o},329.6^{o}\}$
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