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: 16

Answer

$\theta=\{60^o,180^o,300^o\}$

Work Step by Step

$2cos(\theta)+1=sec(\theta)$ $2cos(\theta)+1-sec(\theta)=0\;\;\;\;\;\;\;\;\;\;$ $2cos(\theta)+1-\frac{1}{cos(\theta)}=0\;\;\;\;\;\;\;\;\;\;\;\;$ Multiply each side by $cos(\theta)$ $2cos^2(\theta)+cos(\theta)-1=0\;\;\;\;\;\;\;\;\;\;\;\;$ $cos(\theta)=\frac{-(1)\pm \sqrt{(1)^2-(4.2.(-1))}}{2.2}=-1,\frac{1}{2}$ $cos(\theta)=-1$ $\theta= cos^{-1}(1)$ $\theta=180^o\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ $cos(\theta)=\frac{1}{2}$ $\theta=cos^{-1}(\frac{1}{2})$ We know $ cos(\theta) $ is positive in quadrant $I$ and quadrant $IV$ $\theta=60^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=360^o-60^o=300^o$ $\theta=60^o\;\;\;\;\;\;\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\;\;\;\theta=300^o$ $\theta=\{60^o,180^o,300^o\}$
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