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: 15

Answer

$\theta=\{90^o,210^o,330^o\}$

Work Step by Step

$2sin(\theta)-1=csc(\theta)$ $2sin(\theta)-1-csc(\theta)=0\;\;\;\;\;\;\;\;\;\;$ $2sin(\theta)-1-\frac{1}{sin(\theta)}=0\;\;\;\;\;\;\;\;\;\;\;\;$ multiply each side by $sin(\theta)$ $2sin^2(\theta)-sin(\theta)-1=0\;\;\;\;\;\;\;\;\;\;\;\;$ $sin(\theta)=\frac{-(-1)\pm \sqrt{(-1)^2-(4.2.(-1))}}{2.2}=1,-\frac{1}{2}$ $sin(\theta)=1$ $\theta= sin^{-1}(1)$ $\theta=90^o\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ $sin(\theta)=\frac{-1}{2}$ We know $ sin(\theta) $ is negative in quadrant $III$ and quadrant $IV$ $\theta=180^o+30^o=210^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=360^o-30^o=330^o$ $\theta=210^o\;\;\;\;\;\;\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\;\;\;\theta=330^o$ $\theta=\{90^o,210^o,330^o\}$
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