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: 38

Answer

$\theta=\{120^o,240^o\}$

Work Step by Step

$13cot(\theta)+11sec(\theta)=6sin(\theta)$ $13\frac{cos(\theta)}{sin(\theta)}+11\frac{1}{cos(\theta)}=6sin(\theta)$ $13cos(\theta)+11=6sin^2(\theta)$ $13cos(\theta)+11=6[1-cos^2(\theta)]$ $13cos(\theta)+11=6-6cos^2(\theta)$ $6cos^2(\theta)+13cos(\theta)+5=0$ $cos(\theta)=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(13) \pm \sqrt{(13)^2 - 4 (6)(5)}}{2(6)}=\frac{-5}{3},\frac{-1}{2}$ $cos(\theta)=\frac{-5}{3}$ $\theta=cos^{-1}(\frac{-5}{3})$ $\theta=\varnothing $ $cos(\theta)=\frac{-1}{2}$ $\theta=cos{-1}(\frac{-1}{2})$ We know $cos(\theta)$ is negative in quadrant $II$ and $III$ $\theta=180^o+60^o=240^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=180^o-60^o=120^o$ $\theta=240^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=120^o$ $\theta=\{120^o,240^o\}$
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