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: 45


$\theta=arctan-3+n\pi$ $\theta=arctan~2+n\pi$ where $n$ is an integer.

Work Step by Step

$tan^2\theta+tan~\theta-6=0$ $tan^2\theta+3~tan~\theta-2~tan~\theta-6=0$ $tan~\theta(tan~\theta+3)-2(tan~\theta+3)=0$ $(tan~\theta+3)(tan~\theta-2)=0$ $tan~\theta=-3$ or $tan~\theta=2$ $tan~\theta=-3$ $\theta=arctan-3$ $tan~\theta=2$ $\theta=arctan~2$ The period of $tan~\theta$ is $\pi$. The general solutions are: $\theta=arctan-3+n\pi$ $\theta=arctan~2+n\pi$ where $n$ is an integer.
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