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: 52



Work Step by Step

$2cos^2(\theta)+2sin(\theta)-1=0$ $2[1-sin^2(\theta)]+2sin(\theta)-1=0$ $2-2sin^2(\theta)+2sin(\theta)-1=0$ $2sin^2(\theta)-2sin(\theta)-1=0$ $sin(\theta)=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(2) \pm \sqrt{(2)^2 - 4 (2)(-1)}}{2(2)}=\frac{1+\sqrt{3}}{2},\frac{1-\sqrt{3}}{2}$ $sin(\theta)=\frac{1-\sqrt{3}}{2}=-0.366$ $\theta=sin^{-1}(\frac{1-\sqrt{3}}{2})$ We know $sin(\theta)$ is negative in quadrant $III$ and $IV$ $\theta=180^o-21.5^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=360^o-21.5^o$ $\theta=201.5^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=338.5^o$ $cos(\theta)=\frac{1+\sqrt{3}}{2}=1.366\;\;\;\;$ $\theta=\{201.5^o,338.5^o\}$
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.