Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.3 Trigonometric Equations II - 6.3 Exercises - Page 273: 7

Answer

The solutions of the given equation are $$\{\frac{\pi}{12},\frac{11\pi}{12},\frac{13\pi}{12},\frac{23\pi}{12}\}$$

Work Step by Step

$$\cos2x=\frac{\sqrt3}{2}$$ over interval $[0,2\pi)$ 1) Interval $[0,2\pi)$ can be written as $$0\le x\lt2\pi$$ As a result, for $2x$, the interval would be $$0\le2x\lt4\pi$$ or $$2x\in[0,4\pi)$$ 2) Now consider back the equation $$\cos2x=\frac{\sqrt3}{2}$$ Over the interval $[0,4\pi)$, there are 4 values with $\cos$ equaling $\frac{\sqrt{3}}{2}$, which are $\frac{\pi}{6},\frac{11\pi}{6},\frac{13\pi}{6},\frac{23\pi}{6}$, meaning that $$2x=\{\frac{\pi}{6},\frac{11\pi}{6},\frac{13\pi}{6},\frac{23\pi}{6}\}$$ So $$x=\{\frac{\pi}{12},\frac{11\pi}{12},\frac{13\pi}{12},\frac{23\pi}{12}\}$$
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