Trigonometry 7th Edition

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

Chapter 6 - Section 6.2 - More on Trigonometric Equations - 6.2 Problem Set - Page 333: 59



Work Step by Step

$sin^2(x)-3sin(x)-1=0$ $sin(x)=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(-3) \pm \sqrt{(-3)^2 - 4 (1)(-1)}}{2(1)}=\frac{3+\sqrt{13}}{2},\frac{3-\sqrt{13}}{2}$ $sin(x)=\frac{3-\sqrt{13}}{2}=-0.302$ $x=sin^{-1}(\frac{3-\sqrt{13}}{2})$ We know $sin(x)$ is negative in quadrant $III$ and $IV$ $x=5.9756\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;x=3.4492$ $sin(x)=\frac{3-\sqrt{13}}{2}\;\;\;\;$ $x=\{5.9756,3.4492\}$
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