Functions Modeling Change: A Preparation for Calculus, 5th Edition

Published by Wiley
ISBN 10: 1118583191
ISBN 13: 978-1-11858-319-7

Chapter 7 - Trigonometry and Periodic Functions - 7.8 Inverse Trigonometric Functions - Exercises and Problems for Section 7.8 - Exercises and Problems - Page 314: 16

Answer

$$t=\frac{\pi }{6},\:t=\frac{5\pi }{6}$$

Work Step by Step

Using inverse trigonometric functions to solve the equation, we find: $$\sin \left(t\right)=\frac{1}{2},\:0\le \:t\le \:2\pi \\ t=\frac{\pi }{6}+2\pi n,\:t=\frac{5\pi }{6}+2\pi n \\ t=\frac{\pi }{6},\:t=\frac{5\pi }{6}$$ NOTE: All answers are in radians.
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