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: 54



Work Step by Step

$$\cos2\theta+\cos\theta=0$$ where $0\le\theta\le2\pi$ Apply the Double Angle Formula here for $\cos2\theta$, which states $\cos2\theta=\cos^2\theta-\sin^2\theta$ Yet, we also recall that $\sin^2\theta=1-\cos^2\theta$. Therefore, $$\cos2\theta=\cos^2\theta-(1-\cos^2\theta)=\cos^2\theta-1+\cos^2\theta=2\cos^2\theta-1$$ That means, $$2\cos^2\theta-1+\cos\theta=0$$ $$2\cos^2\theta+\cos\theta-1=0$$ $$(2\cos^2\theta+2\cos\theta)+(-\cos\theta-1)=0$$ $$2\cos\theta(\cos\theta+1)-(\cos\theta+1)=0$$ $$(\cos\theta+1)(2\cos\theta-1)=0$$ $$\cos\theta=-1\hspace{1cm}\text{or}\hspace{1cm}\cos\theta=\frac{1}{2}$$ - For $\cos\theta=-1$: As $\theta\in[0,2\pi]$, there is only 1 point where $\cos\theta=-1$, which is $\theta=\pi$. - For $\cos\theta=\frac{1}{2}$: As $\theta\in[0,2\pi]$, there are 2 points where $\cos\theta=\frac{1}{2}$, which are $\theta=\frac{\pi}{3}$ and $\theta=\frac{5\pi}{3}$. Therefore, overall, the answer to this exercise is $$\theta=\Big[\frac{\pi}{3},\pi,\frac{5\pi}{3}\Big]$$
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