Algebra and Trigonometry 10th Edition

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

Chapter 7 - 7.3 - Solving Trigonometric Equations - 7.3 Exercises - Page 532: 89


$t_1=\frac{1}{8}arctan~\frac{1}{3}\approx0.040~s$ $t_2=\frac{1}{8}arctan~\frac{1}{3}+\frac{1}{8}\pi\approx0.433~s$ $t_3=\frac{1}{8}arctan~\frac{1}{3}+\frac{1}{4}\pi\approx0.826~s$

Work Step by Step

$y=\frac{1}{12}(cos~8t-3~sin~8t)$ $0=\frac{1}{12}(cos~8t-3~sin~8t)$ $0=cos~8t-3~sin~8t$ $3~sin~8t=cos~8t$ $\frac{sin~8t}{cos~8t}=\frac{1}{3}$ $tan~8t=\frac{1}{3}$ $tan$ has a period of $\pi$: $8t=arctan~\frac{1}{3}+n\pi$, where $n$ is an integer. $t=\frac{1}{8}arctan~\frac{1}{3}+\frac{n}{8}\pi$ For $n=0$: $t=\frac{1}{8}arctan~\frac{1}{3}\approx0.040$ For $n=1$: $t=\frac{1}{8}arctan~\frac{1}{3}+\frac{1}{8}\pi\approx0.433$ For $n=2$: $t=\frac{1}{8}arctan~\frac{1}{3}+\frac{2}{8}\pi\approx0.826$
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