Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 7 - 7.3 - Solving Trigonometric Equations - 7.3 Exercises - Page 530: 17


$x=\frac{\pi}{6}+n\pi$ and $x=\frac{5\pi}{6}+n\pi$, where $n$ is an integer.

Work Step by Step

$4~cos^2x-1=0$ $4~cos^2x=1$ $cos^2x=\frac{1}{4}$ $cos~x=±\frac{1}{2}$ The period of $cos~x$ is $2\pi$. The solutions in the interval: $[0,2\pi)$ are: $x=\frac{\pi}{6}$, $x=\frac{5\pi}{6}$, $x=\frac{7\pi}{6}$ and $x=\frac{11\pi}{6}$ Now, add multiples of $2\pi$ to each of the solutions: $x=\frac{\pi}{6}+2n\pi$, $x=\frac{5\pi}{6}+2n\pi$, $x=\frac{7\pi}{6}+2n\pi$ and $x=\frac{11\pi}{6}+2n\pi$ But, notice that $x=\frac{7\pi}{6}=\frac{\pi}{6}+\pi$ and $x=\frac{11\pi}{6}=\frac{5\pi}{6}+\pi$ We can rewrite the solutions as: $x=\frac{\pi}{6}+n\pi$ and $x=\frac{5\pi}{6}+n\pi$, where $n$ is an integer.
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