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 274: 65b

Answer

From the graph, we can see that $P \leq 0$ on the interval $[0.001638,~0.003549]$

Work Step by Step

$P=0~~$ when $~~2\pi(261.63)t+\frac{\pi}{7} = \pi~n$, where $n$ is an integer. $t = \frac{7n-1}{(2)(261.63)(7)}$ $t = \frac{7n-1}{3662.82}$ We can find the first positive value of $t$ such that $P=0$: $t = \frac{(7)(1)-1}{3662.82}$ $t = 0.001638~s$ We can find the second positive value of $t$ such that $P=0$: $t = \frac{(7)(2)-1}{3662.82}$ $t = 0.003549~s$ From the graph, we can see that $P \leq 0$ on the interval $[0.001638,~0.003549]$
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