University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.3 - Trigonometric Functions - Exercises - Page 28: 53



Work Step by Step

$$\sin2\theta-\cos\theta=0$$ where $0\le\theta\le2\pi$ Apply the Double Angle Formula here for $\sin2\theta$, which states $\sin2\theta=2\sin\theta\cos\theta$ That means, $$2\sin\theta\cos\theta-\cos\theta=0$$ $$\cos\theta(2\sin\theta-1)=0$$ $$\cos\theta=0\hspace{1cm}\text{or}\hspace{1cm}\sin\theta=\frac{1}{2}$$ - For $\cos\theta=0$: As $\theta\in[0,2\pi]$, there are 2 points where $\cos\theta=0$, which are $\theta=\frac{\pi}{2}$ and $\theta=\frac{3\pi}{2}$. - For $\sin\theta=\frac{1}{2}$: As $\theta\in[0,2\pi]$, there are also 2 points where $\sin\theta=\frac{1}{2}$, which are $\theta=\frac{\pi}{6}$ and $\theta=\frac{5\pi}{6}$. Therefore, overall, the answer to this exercise is $$\theta=\Big[\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6},\frac{3\pi}{2}\Big]$$
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