Algebra and Trigonometry 10th Edition

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

Chapter 7 - 7.3 - Solving Trigonometric Equations - 7.3 Exercises - Page 531: 65


$x=arctan(\frac{1}{3})+n\pi$ and $x=arctan(-\frac{1}{3})+n\pi$, where $n$ is an integer.

Work Step by Step

$cot^2x-9=0$ $cot^2x=9$ $cot~x=±3$ $tan~x=±\frac{1}{3}$ $x=arctan(\frac{1}{3})$ $x=arctan(-\frac{1}{3})$ The period of $tan~x$ is $\pi$. So, add multiples of $\pi$ to each solution to find the general solution: $x=arctan(\frac{1}{3})+n\pi$ and $x=arctan(-\frac{1}{3})+n\pi$, where $n$ is an integer.
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