Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 6 - Section 6.1 - Solving Trigonometric Equations - 6.1 Problem Set - Page 325: 8


$(a)$ $\theta=150^{o}+360^{o}k $ or $\theta=210^{o}+360^{o}k$ $(b)$ $150^{o}$ and $210^{o}$

Work Step by Step

The first task is to isolate the trigonometric function on one side: Subtract $\sqrt{3}$ to both sides... $2\cos\theta=-\sqrt{3}\qquad $ ... divide with 2 $\displaystyle \cos\theta=- \frac{\sqrt{3}}{2}$ Now, we find a reference angle. From the table of characteristic angles, we know that $\displaystyle \cos 30^{o}=\frac{\sqrt{3}}{2}.$ Next, we know that cosine is negative in quadrants I and III, so angles that satisfy the equation are $180^{o}-30^{o}=150^{o}$ and $180^{o}+30^{o}=210^{o}$ Finally, to each individual solution, add multiples of $360^{o}$ to cover all solutions: $(a)$ $\theta=150^{o}+360^{o}k $ or $\theta=210^{o}+360^{o}k $ $(b)$ The solutions within the interval $ 0^{o}\leq\theta \lt 360^{o}:$ $150^{o}$ and $210^{o}.$
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