Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 1 - Functions - 1.4 Trigonometric Functions and Their Inverse - 1.4 Exercises - Page 48: 39

Answer

$\boxed{\theta= \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}}$

Work Step by Step

$sin^2(\theta) = \frac{1}{4}$ which means $sin(\theta) = +\frac{1}{2}$ or $sin(\theta) = -\frac{1}{2}$. For the first case, $sin(\theta) = +\frac{1}{2}$, $\theta = \frac{\pi}{6}, \frac{5\pi}{6}$. For the second case, $sin(\theta) = -\frac{1}{2}$, $\theta = \frac{7\pi}{6}, \frac{11\pi}{6}$. Overall, the solutions for $\theta$ are: $\frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$
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