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: 59


$x=arctan~3+n\pi$ and $x=arctan(-4)+n\pi$, where $n$ is an integer.

Work Step by Step

Use: $tan~x=4~tan~x-3~tan~x$ $tan^2x+tan~x-12=0~~$ $tan^2x+4~tan~x-3~tan~x-12=0$ $tan~x(tan~x+4)-3(tan~x+4)=0$ $(tan~x-3)(tan~x+4)=0$ $tan~x-3=0$ $tan~x=3$ $x=arctan~3$ $tan~x+4=0$ $tan~x=-4$ $x=arctan(-4)$ The period of $tan~x$ is $\pi$. So, add multiples of $\pi$ to each solution to find the general solution: $x=arctan~3+n\pi$ and $x=arctan(-4)+n\pi$, where $n$ is an integer.
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