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: 39

Answer

$\theta=\{41.8^o,48.6^o,131.4^o,138.2^o\}$

Work Step by Step

$18sec^2(\theta)-17tan(\theta)sec(\theta)-12=0$ $18\frac{1}{cos^2(\theta)}-17\frac{sin(\theta)}{cos(\theta)}.\frac{1}{cos(\theta)}-12=0$ $18-17sin(\theta)-12cos^2(\theta)=0$ $18-17sin(\theta)-12[1-sin^2(\theta)]=0$ $18-17sin(\theta)-12+12sin^2(\theta)=0$ $12sin^2(\theta)-17sin(\theta)+6=0$ $sin(\theta)=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(-17) \pm \sqrt{(-17)^2 - 4 (12)(6)}}{2(12)}=\frac{3}{4},\frac{2}{3}$ $sin(\theta)=\frac{3}{4}$ $\theta=sin^{-1}(\frac{3}{4})$ We know $sin(\theta)$ is positive in quadrant $I$ and $II$ $\theta=48.6^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=180^o-48.6^o=131.4^o$ $\theta=48.6^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=131.4^o$ $sin(\theta)=\frac{2}{3}$ $\theta=sin{-1}(\frac{2}{3})$ We know $sin(\theta)$ is positive in quadrant $I$ and $II$ $\theta=41.8^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=180^o-41.8^o=138.2^o$ $\theta=41.8^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=138.2^o$ $\theta=\{41.8^o,48.6^o,131.4^o,138.2^o\}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.