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: 54



Work Step by Step

$sin^2(\theta)-cos(\theta)=0$ $1-cos^2(\theta)-cos(\theta)=0$ $cos^2(\theta)+cos(\theta)-1=0$ $cos(\theta)=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(1) \pm \sqrt{(1)^2 - 4 (1)(-1)}}{2(1)}=\frac{-1+\sqrt{5}}{2},\frac{-1-\sqrt{5}}{2}$ $cos(\theta)=\frac{-1+\sqrt{5}}{2}=0.61$ $\theta=cos^{-1}(\frac{-1+\sqrt{5}}{2})$ We know $cos(\theta)$ is positive in quadrant $I$ and $IV$ $\theta=51.8^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=360^o-51.8^o$ $\theta=51.8^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=308.2^o$ $cos(\theta)=\frac{-1-\sqrt{5}}{2}=1.6\;\;\;\;$ $\theta=\{51.8^o,308.2^o\}$
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