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: 67a



Work Step by Step

$$V=\cos2\pi t$$ where $t\in[0,\frac{1}{2}]$ As $V=0$, we would have the equation: $$\cos2\pi t=0$$ We would take the interval $[0,2\pi)$ here. Over the interval $[0,2\pi)$, there are 2 values of $2\pi t$ where $\cos2\pi t=0$, which are $\frac{\pi}{2}$ and $\frac{3\pi}{2}$. That means $$2\pi t=\frac{\pi}{2}\hspace{1cm}\text{or}\hspace{1cm}2\pi t=\frac{3\pi}{2}$$ $$t=\frac{\pi}{2\times2\pi}\hspace{1cm}\text{or}\hspace{1cm}t=\frac{3\pi}{2\times2\pi}$$ $$t=\frac{1}{4}\hspace{1cm}\text{or}\hspace{1cm}t=\frac{3}{4}$$ $t=\frac{3}{4}$ lies out of the defined interval for $t$: $t\in[0,\frac{1}{2}]$, so we need to eliminate $t=\frac{3}{4}$ That means $t=\frac{1}{4}$ is the result of this question.
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