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: 32


$u=\frac{\pi}{6}+n\pi$ and $u=\frac{5\pi}{6}+n\pi$, where $n$ is an integer.

Work Step by Step

$4~tan^2u-1=tan^2u$ $3~tan^2u=1$ $tan^2u=\frac{1}{3}$ $tan~u=±\frac{1}{\sqrt 3}=±\frac{\sqrt 3}{3}$ The period of $tan~u$ is $\pi$. The solutions in the interval $[0,\pi)$ are: $u=\frac{\pi}{6}$ and $u=\frac{5\pi}{6}$ Now, add multiples of $\pi$ to the solutions: $u=\frac{\pi}{6}+n\pi$ and $u=\frac{5\pi}{6}+n\pi$, where $n$ is an integer.
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