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: 8

Answer

$\theta=150^o\;\;\;\;\;\;\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\;\;\;\theta=210^o$

Work Step by Step

$2\sqrt{3}sec(\Theta )+7=3$ $2\sqrt{3}sec(\Theta )=3-7=-4\;\;\;\;\;\;\;\;\;\;$ subtract $ 7 $ from each side. $sec(\Theta )=\frac{-4}{2\sqrt{3}} \;\;\;\;\;\;\;\;\;\;\;$ divide each side by $2\sqrt{3}$ $\frac{1}{cos(\Theta )}=\frac{-2}{\sqrt{3}}$ $cos(\theta)=\frac{\sqrt{3}}{-2}$ $\theta= cos^{-1}(\frac{\sqrt{3}}{-2})$ We know $ cos(\theta) $ is negative in quadrant $II$ and quadrant $III$ $\theta=180^o-30^o=150^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=180^o+30^o=210^o$ $\theta=150^o\;\;\;\;\;\;\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\;\;\;\theta=210^o$
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