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: 46


$x=arctan-5+n\pi$ $x=-\frac{\pi}{4}+n\pi$ where $n$ is an integer.

Work Step by Step

$sec^2x+6~tan~x+4=0$ $1+tan^2x+6~tan~x+4=0$ $tan^2x+6~tan~x+5=0$ $tan^2x+tan~x+5~tan~x+5=0$ $tan~x(tan~x+1)+5(tan~x+1)=0$ $(tan~x+5)(tan~x+1)=0$ $tan~x=-5$ or $tan~x=-1$ $tan~x=-5$ $x=arctan-5$ $tan~x=-1$ $x=-\frac{\pi}{4}$ The period of $tan~x$ is $\pi$. The general solutions are: $x=arctan-5+n\pi$ $x=-\frac{\pi}{4}+n\pi$ where $n$ is an integer.
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