Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 6 - Section 6.3 - Trigonometric Equations Involving Multiple Angles - 6.3 Problem Set - Page 340: 23

Answer

$\theta=\{112.5^o\;,157.5^o,292.5^o,337.5^o\},\;\;\;\;\;\;\;\;\;\;$

Work Step by Step

$sin(2x)=\frac{-\sqrt{2}}{2}$ $2x=sin^{-1}(\frac{-\sqrt{2}}{2})$ We know $sin(x)$ is negative in quardent $III$ and quardent $IV$ The period of the sine function is $360^o$ $2x=180^o+45^o\;\;\;\;\;\;\;or\;\;\;\;2x=360^o+225^o\;\;\;or\;\;\;\;\;\;\;2\theta=360^o-45^o\;\;\;\;\;\;\;or\;\;\;\;\;\;2x=360^o+315^o$ $2x=225^o\;\;\;\;\;\;\;or\;\;\;\;2x=585^o\;\;\;or\;\;\;\;\;\;\;2x=315^o\;\;\;\;\;\;\;or\;\;\;\;\;\;2x=675^o $ $x=112.5^o\;\;\;\;\;\;\;or\;\;\;\;x=157.5^o\;\;\;\;\;or\;\;\;\;\;\;\;x=292.5^o\;\;\;\;\;\;\;or\;\;\;\;\;\;\;x=337.5^o $ $\theta=\{112.5^o\;,157.5^o,292.5^o,337.5^o\}\;\;\;\;\;\;\;\;\;\;$
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