Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 7 - Review Exercises - Page 552: 29



Work Step by Step

$3\sqrt 3~tan~u=3$ $tan~u=\frac{3}{3\sqrt 3}=\frac{\sqrt 3~·\sqrt 3}{3\sqrt 3}$ $tan~u=\frac{\sqrt 3}{3}$ The period of $tan~u$ is $\pi$. The solution in the interval: $[0,\pi)$ is: $u=\frac{\pi}{6}$ Now, add multiples of $\pi$ to the solution: $u=\frac{\pi}{6}+n\pi$
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