Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 6 - Section 6.3 - Trigonometric Equations Involving Multiple Angles - 6.3 Problem Set - Page 339: 8

Answer

$\theta=\{22.5^o,112.5^o,202.5^o,292.5^o\}$

Work Step by Step

$cot(2\theta)=1$ $\frac{1}{tan(2\theta)}=1$ $tan(2\theta)=1$ $2\theta=tan^{-1}(1)$ We know $tan(\theta)$ is positive in quardent $I$ and quardent $III$ The period of the tangent function is $360^o$ $2\theta=45^o\;\;\;\;or\;\;\;\;\;\;2\theta=360^o+45^o\;\;\;\;or\;\;\;\;\;\;2\theta=180^o+45^o\;\;\;\;or\;\;\;\;\;\;2\theta=360^o+135^o$ $\theta=22.5^o\;\;\;\;or\;\;\;\;\;\;\theta=202.5^o\;\;\;\;or\;\;\;\;\;\;\theta=112.5^o\;\;\;\;or\;\;\;\;\;\;\theta=292.5^o\;\;\;$ $\theta=\{22.5^o,112.5^o,202.5^o,292.5^o\}$
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