Trigonometry (10th Edition)

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

Chapter 6 - Test - Page 287: 13

Answer

$\left\{\frac{\pi}{12}, \frac{7\pi}{12}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{17\pi}{12} \frac{23\pi}{12}\right\}$

Work Step by Step

Add $1$ to both sides: $\sqrt2 \cdot \cos{(3x)}=1$ Divide $\sqrt2$ to both sides: $\cos{(3x)}=\frac{1}{\sqrt2} \\\cos{(3x)}=\frac{\sqrt2}{2} \\3x=\cos^{-1}{(\frac{\sqrt2}{2})}$ Use a scientific calculator's inverse cosine function to obtain: $3x=\frac{\pi}{4} \text{ or } \frac{7\pi}{4} \text{ or } \frac{9\pi}{4} \text{ or } \frac{15\pi}{4} \text{ or } \frac{17\pi}{4} \text{ or } \frac{23\pi}{4} \\x=\frac{\pi}{12} \text{ or } \frac{7\pi}{12} \text{ or } \frac{9\pi}{12} \text{ or } \frac{15\pi}{12} \text{ or } \frac{17\pi}{12} \text{ or } \frac{23\pi}{12} \\x=\frac{\pi}{12} \text{ or } \frac{7\pi}{12} \text{ or } \frac{3\pi}{4} \text{ or } \frac{5\pi}{4} \text{ or } \frac{17\pi}{12} \text{ or } \frac{23\pi}{12}$
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