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: 18


$x=\frac{\pi}{4}+n·\frac{\pi}{2}$, where $n$ is an integer.

Work Step by Step

$2-4~sin^2x=0$ $2=4~sin^2x$ $sin^2x=\frac{1}{2}$ $sin~x=±\sqrt {\frac{1}{2}}=±\frac{\sqrt 2}{2}$ The period of $sin~x$ is $2\pi$. The solutions in the interval: $[0,2\pi)$ are: $x=\frac{\pi}{4}$, $x=\frac{3\pi}{4}$, $x=\frac{5\pi}{4}$, $x=\frac{7\pi}{4}$ Notice that: $x=\frac{3\pi}{4}=\frac{\pi}{4}+1·\frac{\pi}{2}$, $x=\frac{5\pi}{4}=\frac{\pi}{4}+2·\frac{\pi}{2}$, $x=\frac{7\pi}{4}=\frac{\pi}{4}+3·\frac{\pi}{2}$ Now, add multiples of $2\pi$ to each of the solutions: $x=\frac{\pi}{4}+2n\pi$ $x=\frac{3\pi}{4}=\frac{\pi}{4}+\frac{\pi}{2}+2n\pi$, $x=\frac{5\pi}{4}=\frac{\pi}{4}+2·\frac{\pi}{2}+2n\pi$, $x=\frac{7\pi}{4}=\frac{\pi}{4}+3·\frac{\pi}{2}+2n\pi$ We can rewrite all the solutions as: $x=\frac{\pi}{4}+n·\frac{\pi}{2}$, where $n$ is an integer.
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