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: 37



Work Step by Step

$6cos(\theta)+7tan(\theta)=sec(\theta)$ $6cos(\theta)+7\frac{sin(\theta)}{cos(\theta)}=\frac{1}{cos(\theta)}$ $6cos^2(\theta)+7sin(\theta)=1$ $6[1-sin^2(\theta)]+7sin(\theta)-1=0$ $6-6sin^2(\theta)+7sin(\theta)-1=0$ $6sin^2(\theta)-7sin(\theta)-5=0$ $sin(\theta)=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(-7) \pm \sqrt{(-7)^2 - 4 (6)(-5)}}{2(6)}=\frac{5}{3},\frac{-1}{2}$ $sin(\theta)=\frac{5}{3}$ $\theta=sin^{-1}(\frac{5}{3})$ $\theta=\varnothing $ $sin(\theta)=\frac{-1}{2}$ $\theta=sin^{-1}(\frac{-1}{2})$ We know $sin(\theta)$ is negative in quadrant $III$ and $IV$ $\theta=180^o+30^o=120^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=360^o-30^o=330^o$ $\theta=210^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=330^o$ $\theta=\{210^o,330^o\}$
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