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: 10



Work Step by Step

$4cos(\theta)-3sec(\theta)=0$ $4cos(\theta)-3\frac{1}{cos(\theta)}=0\;\;\;\;\;\;\;\;\;\;$ multiply each side by $cos(\theta)$. $4cos^2(\theta)-3=0$ $4cos^2(\Theta )=3\;\;\;\;\;\;\;\;\;\;$ subtract $ -3 $ from each side. $cos^2(\Theta )=\frac{3}{4} \;\;\;\;\;\;\;\;\;\;\;$ divide each side by $4$ $cos(\Theta )=\pm \sqrt{\frac{3}{4}}$ $cos(\Theta )=\pm \frac{\sqrt{3}}{2}$ $\theta= cos^{-1}(\frac{\sqrt{3}}{2})$ We know $ cos(\theta) $ is positive in quadrant $I$ and quadrant $IV$ $\theta=30^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=360^o-30^o=330^o$ $\theta=30^o\;\;\;\;\;\;\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\;\;\;\theta=330^o$ $\theta= cos^{-1}(\frac{-\sqrt{3}}{2})$ We know $ cos(\theta) $ is negative in quadrant $II$ and quadrant $III$ $\theta=180^o+30^o=210^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=180^o-30^o=150^o$ $\theta=210^o\;\;\;\;\;\;\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\;\;\;\theta=150^o$ $\theta=\{30^o,150^o,210^o,330^o\}$
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