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: 72


$\theta\in\{0^{o},120^{o}, 240^{o}\}$

Work Step by Step

$\sin 3\theta=0$ If $0^{o} \leq \theta < 360^{o}$ then $0^{o} \leq 3\theta < 1080^{0}$ So, $3\theta\in[0,1080^{o})$, 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 $0^{o}$, but also $0^{o}+360^{o}=360^{o}$, and $0^{o}+720^{o}=720^{0}$ So, $3\theta\in\{ 0^{o},360^{o},720^{o}\}$, and, since $(3\theta)\div 3=\theta,$ after dividing each corresponding angle with 3, $\theta\in\{0^{o},120^{o}, 240^{o}\}$
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