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: 71


$\theta\in\{90^{o},210^{o}, 330^{o}\}$

Work Step by Step

$\sin 3\theta=-1$ If $0^{o} \leq \theta < 360^{o}$ then $0^{o} \leq 3\theta < 1080^{0}$ So, $3\theta\in[0,1080^{0})$, meaning that if $ 3\theta$ (a solution) is any angle from $[0^{o},360^{o})$, then $(3\theta+360^{o})$ and $(3\theta+720^{o})$ are also solutions. Using the unit circle,$ 3\theta$ can be $270^{o}$, but also $270^{o}+360^{o}=630^{o}$, and $270^{o}+720^{o}=990^{0}$ So, $3\theta\in\{ 270^{o},630^{o},990^{o}\}$, and, since $(3\theta)\div 3=\theta,$ after dividing each corresponding angle with 3, $\theta\in\{90^{o},210^{o}, 330^{o}\}$
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