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 333: 43

Answer

$\theta=225^o+360k\;\;\;\;\;\;\;\;\;\;\;\;\;,\;\;\;\;\;\;\;\;\theta=315^o+360k$

Work Step by Step

$\sqrt{2}csc(\theta)+5=3$ $\frac{\sqrt{2}}{sin(\theta)}=3-5$ $\sqrt{2}=-2sin(\theta)$ $sin(\theta)=\frac{-\sqrt{2}}{2}$ $\theta=sin^{-1}(\frac{-\sqrt{2}}{2})$ We know $sin(\theta)$ is negative in quadrant $III$ and $IV$ $\theta=180^o+45^o=225^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=360^o-45^o=315^o$ $\theta=225^o+360k\;\;\;\;\;\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=315^o+360k\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;where \;\;\;\;\;\;\;\;\;\;\;\;k=\{0,1,2,3,......\}$
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