Algebra and Trigonometry 10th Edition

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

Chapter 7 - 7.4 - Sum and Difference Equations - 7.4 Exercises - Page 539: 71


$x=\dfrac{5\pi}{4}, \dfrac{ 7\pi}{4}$

Work Step by Step

We have $ \cos (x +\dfrac{\pi}{4}) =\cos x \cos \dfrac{\pi}{4}-\sin x \sin \dfrac{\pi}{4}=\dfrac{\sqrt 2}{2} \cos x -\dfrac{\sqrt 2}{2} \sin x$ $ \cos (x -\dfrac{\pi}{4}) =\cos x \cos \dfrac{\pi}{4}+\sin x \sin \dfrac{\pi}{4}=\dfrac{\sqrt 2}{2} \cos x +\dfrac{\sqrt 2}{2} \sin x$ After simplifying, we get $\dfrac{\sqrt 2}{2} \cos x -\dfrac{\sqrt 2}{2} \sin x - \dfrac{\sqrt 2}{2} \cos x -\dfrac{\sqrt 2}{2} \sin x= -\sqrt 2 \sin x=1$ or, $\sin x =\dfrac{-1}{\sqrt 2}$ or, $\sin x=\dfrac{-\sqrt 2}{2}$ This yields: $x=\dfrac{5\pi}{4}, \dfrac{ 7\pi}{4}$
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