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 332: 31

Answer

$\theta=\{60^o,180^o\}$

Work Step by Step

$\sqrt{3}sin(\theta)-cos(\theta)=1$ $\sqrt{3}sin(\theta)=1+cos(\theta)$ $(\sqrt{3}sin(\theta))^2=(1+cos(\theta))^2\;\;\;\;\;\;\;\;\;\;$ $3sin^2(\theta)=1+2cos(\theta)+cos^2(\theta)\;\;\;\;\;\;\;\;\;\;\;\;$ $3-3cos^2(\theta)=1+2cos(\theta)+cos^2(\theta)$ $4cos^2(\theta)+2cos(\theta)-2=0$ $2cos^2(\theta)+cos(\theta)-1=0$ $cos(\theta)=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(1) \pm \sqrt{(1)^2-4(2)(-1)}}{2(2)}=-1,\frac{1}{2}$ $cos(\theta)=-1$ $\theta= cos^{-1}(-1)$ $\theta=180^o\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ $cos(\theta)=\frac{1}{2}$ $\theta=cos^{-1}(\frac{1}{2})$ We know $ cos(\theta) $ is positive in quadrant $I$ and quadrant $IV$ $\theta=60^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=360^o-60^o$ $\theta=60^o\;\;\;\;\;\;\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\;\;\;\theta=300^o$ Refuse $\theta=\{60^o,180^o\}$
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