Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.2 Trigonometric Equations I - 6.2 Exercises - Page 273: 27

Answer

$\theta = 30^{\circ}, 240^{\circ}, 240^{\circ}, 300^{\circ}$

Work Step by Step

$(\cot \theta- \sqrt 3)(2\sin \theta+ \sqrt 3) = 0$ $\cot \theta - \sqrt 3= 0$ $\cot \theta = \sqrt 3$ $\frac{1}{\tan \theta} = \sqrt 3$ $1 = \tan \theta \sqrt 3$ $\tan \theta = \frac{1}{\sqrt 3}$ $\tan \theta = \frac{\sqrt 3}{3}$ $\theta = 30^{\circ}, 240^{\circ}$ $2\sin \theta + \sqrt 3 = 0$ $2\sin \theta = - \sqrt 3$ $\sin \theta = - \frac{\sqrt 3}{2}$ $\theta = 240^{\circ}, 300^{\circ}$ Therefore, $\theta = 30^{\circ}, 240^{\circ}, 240^{\circ}, 300^{\circ}$
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