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: 23

Answer

$x=\{\frac{\pi}{3},\frac{5\pi}{3}\}$

Work Step by Step

$4sin^2(x)+4cos(x)-5=0$ $4-4cos^2(x)+4cos(x)-5=0\;\;\;\;\;\;\;\;\;\;$ $4cos^2(x)-4cos(x)+1=0\;\;\;\;\;\;\;\;\;\;\;\;$ $cos(x)=\frac{-(-4)\pm \sqrt{(-4)^2-(4.4.(1))}}{2.4}=\frac{1}{2}$ $cos(x)=\frac{1}{2}$ $x=cos^{-1}(\frac{1}{2})$ We know $ cos(x) $ is positive in quadrant $I$ and quadrant $IV$ $x=\frac{\pi}{3}\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;x=2\pi-\frac{\pi}{3}$ $x=\frac{\pi}{3}\;\;\;\;\;\;\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\;\;\;x=\frac{5\pi}{3}$ $x=\{\frac{\pi}{3},\frac{5\pi}{3}\}$
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