Trigonometry 7th Edition

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

Chapter 6 - Section 6.2 - More on Trigonometric Equations - 6.2 Problem Set - Page 332: 7

Answer

$\theta=225^o\;\;\;\;\;\;\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\;\;\;\theta=315^o$

Work Step by Step

$\sqrt{2}csc(\Theta )+5=3$ $\sqrt{2}csc(\Theta )=3-5=-2\;\;\;\;\;\;\;\;\;\;$ subtract $ 5 $ from each side. $csc(\Theta )=\frac{-2}{\sqrt{2}} \;\;\;\;\;\;\;\;\;\;\;$ divide each side by $\sqrt{2}$ $\frac{1}{sin(\Theta )}=\frac{-2}{\sqrt{2}}$ $sin(\theta)=\frac{\sqrt{2}}{-2}$ $\theta= sin^{-1}(\frac{\sqrt{2}}{-2})$ We know $ sin(\theta) $ is negative in quadrant 3 and quadrant 4 $\theta=45^o+180^o=225^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=360^o-45^o=315^o$ $\theta=225^o\;\;\;\;\;\;\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\;\;\;\theta=315^o$
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