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: 47

Answer

$\{180^{o}+k\cdot 360^{o}, $ k any integer$\}$

Work Step by Step

$\cos\theta=-1$ In the interval $[0,360^{o})$, a solution is $\theta=180^{o}$ Since the period for cosine is $360^{o}$, then all angles $180^{o}+k\cdot 360^{o}, $ k any integer, are also solutions. Solution set: $\{180^{o}+k\cdot 360^{o}, $ k any integer$\}$
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