Algebra and Trigonometry 10th Edition

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

Chapter 7 - Chapter Test - Page 554: 14


$\alpha = \dfrac{\pi}{6},\dfrac{\pi}{2}, \dfrac{5\pi}{6},\dfrac{3\pi}{2}$

Work Step by Step

$2 cos \alpha \sin \alpha -\cos \alpha= 0$ We can simplify the equation as follows: $ cos \alpha (2 \sin \alpha - 1)= 0$ $cos \alpha =0 $ and $\sin \alpha= \dfrac{1}{2}$ This gives 4 possible solutions: $\alpha = \dfrac{\pi}{6},\dfrac{\pi}{2}, \dfrac{5\pi}{6},\dfrac{3\pi}{2}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.