Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 6 - Test - Page 287: 17

Answer

The smallest non-negative solution is $x = \frac{\pi}{2}$ In general, the non-negative solutions have the form: $x = \frac{\pi}{2} + 2\pi~n$, where $n$ is a non-negative integer

Work Step by Step

$csc~x-cot~x = 1$ $\frac{1}{sin~x}-\frac{cos~x}{sin~x} = 1$ $1-cos~x = sin~x$ $1-2~cos~x+cos^2~x = sin^2~x$ $sin^2~x+cos^2~x-2~cos~x+cos^2~x = sin^2~x$ $2~cos^2~x-2~cos~x = 0$ $2~cos^2~x = 2~cos~x$ $cos^2~x = cos~x$ Then $cos~x = 0$ or $cos~x = 1$ If $cos~x = 1$, then $x = 0$. However, then $csc~x$ is undefined in the original equation. Thus $x = 0$ is not a valid solution. Therefore, $cos~x = 0$. Then in the original equation: $csc~x-cot~x = 1$ $csc~x-0 = 1$ $csc~x = 1$ $sin~x = 1$ $x = arcsin(1)$ $x = \frac{\pi}{2}$ The smallest non-negative solution is $x = \frac{\pi}{2}$ In general, the non-negative solutions have the form: $x = \frac{\pi}{2} + 2\pi~n$, where $n$ is a non-negative integer
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