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: 49

Answer

$\displaystyle \{\frac{\pi}{3}+2k\pi, \frac{2\pi}{3}+2k\pi, $ (k any integer)$\}$

Work Step by Step

3$\csc x=2\sqrt{3}\qquad/\div 3$ $\displaystyle \csc x=\frac{2\sqrt{3}}{3}$ so $\displaystyle \sin x=\frac{3}{2\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{3\sqrt{3}}{2(3)}=\frac{\sqrt{3}}{2}$ From the unit circle, $\displaystyle \sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$ and $\displaystyle \sin\frac{2\pi}{3}=\frac{\sqrt{3}}{2}$ so, since sine has a period of $2\pi,$ the solutions will have form $\displaystyle \frac{\pi}{3}+2k\pi, $ k any integer or $\displaystyle \frac{2\pi}{3}+2k\pi, $ k any integer Solution set: $\displaystyle \{\frac{\pi}{3}+2k\pi, \frac{2\pi}{3}+2k\pi, $ (k any integer)$\}$
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