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: 72


$x=\dfrac{\pi}{6}, \dfrac{5\pi}{6}$

Work Step by Step

We have $(\sin x \cos \dfrac{\pi}{6}+\cos x \sin \dfrac{\pi}{6})-(\sin x \cos \dfrac{7\pi}{6}-\cos x \sin \dfrac{7\pi}{6}) =\dfrac{\sqrt 3}{2}$ $ (\dfrac{\sqrt 3}{2} \sin x +\dfrac{1}{2} \cos x)-(-\dfrac{\sqrt 3}{2} \sin x -(-\dfrac{1}{2}) \cos x)=\dfrac{\sqrt 3}{2} $ After simplifying, we get $\dfrac{\sqrt 3}{2} \sin x +\dfrac{1}{2} \cos x + \dfrac{\sqrt 3}{2} \sin x -\dfrac{1}{2} \cos x=\dfrac{\sqrt 3}{2}$ or, $ \sqrt 3 \sin x =\dfrac{\sqrt 3}{2}$ or, $\sin x=\dfrac{1}{2}$ This yields: $x=\dfrac{\pi}{6}, \dfrac{5\pi}{6}$
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