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: 65a



Work Step by Step

The question asks to find the values of $t$ for which $P=0$ over $[0,0.005]$. So we would use the formula of $P$ where $x=t$, which is $$P(t)=0.004\sin[2\pi(261.63)t+\frac{\pi}{7}]$$ where $P=0$, which means $$0.004\sin[2\pi(261.63)t+\frac{\pi}{7}]=0$$ The interval is $[0,0.005]$ Now we solve the equation: $$0.004\sin[2\pi(261.63)t+\frac{\pi}{7}]=0$$ $$\sin[2\pi(261.63)t+\frac{\pi}{7}]=0\hspace{1cm}(1)$$ Over the interval $[0,0.005]$, there is one value which has sine equaling $0$, which is $0$ ($\sin0=0$) Apply it to $(1)$: $$2\pi(261.63)t+\frac{\pi}{7}=0$$ $$2\pi(261.63)t=-\frac{\pi}{7}$$ $$261.63t=-\frac{\pi}{7\times2\pi}=-\frac{1}{14}$$ $$t=-\frac{1}{14\times261.63}$$ $$t\approx-2.73\times10^{-4}$$ That is the value of $t$ needed to find.
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