Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 7 - Analytic Trigonometry - 7.3 Trigonometric Equations - 7.3 Assess Your Understanding - Page 466: 28



Work Step by Step

$\tan(\frac{\theta}{2})=\sqrt3$ $\frac{\theta}{2}=\frac{1}{3}\pi+k\pi$ (since the period of tangent is $\pi$) $\theta=\frac{2}{3}\pi+2k\pi$ Those solutions where $0\le\theta\lt2\pi$ are only: $\theta=\frac{2}{3}\pi+2(0)\pi=\frac{2}{3}\pi$
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