Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 14 Trigonometric Graphs, Identities, and Equations - 14.5 Write Trigonometric Functions and Models - 14.5 Exercises - Quiz for Lessons 14.3-14.5 - Page 945: 8


$x =2 n \pi \pm \dfrac{\pi}{2}$ and $x =n \pi + (-1)^n \dfrac{\pi}{4}$

Work Step by Step

Re-arrange the given equation as: $\sqrt 2 \cos x(\sin x-\dfrac{1}{\sqrt 2})=0$ When $\sqrt 2 \cos x=0$ Therefore, the general solution of $\cos x$ is: $x =2 n \pi \pm \dfrac{\pi}{2}$ when $\sin x=\dfrac{1}{\sqrt 2}$ Therefore, the general solution of $\cos x$ is: $x =n \pi + (-1)^n \dfrac{\pi}{4}$
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