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: 70


$x=arcsin\frac{1}{5}+2n\pi$, where $n$ is an integer.

Work Step by Step

$csc^2x-5~csc~x=0$ $csc~x(csc~x-5)=0$ $csc~x=0$. But, there is no $x$ such that $csc~x=0$. $csc~x-5=0$ $csc~x=5$ $\frac{1}{sin~x}=5$ $sin~x=\frac{1}{5}$ $x=arcsin\frac{1}{5}$ The period of $sin~x$ is $2\pi$. So, add multiples of $2\pi$ to each solution to find the general solution: $x=arcsin\frac{1}{5}+2n\pi$, where $n$ is an integer.
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