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: 58

Answer

$x=\{2.4981,4.7124\}$

Work Step by Step

$2cos(x)+sin(x)+1=0$ $[sin(x)+1]^2=[-2cos(x)]^2$ $sin^2(x)+2sin(x)+1=4cos^2(x)$ $sin^2(x)+2sin(x)+1-4cos^2(x)=0$ $sin^2(x)+2sin(x)+1-4[1-sin^2(x)]=0$ $sin^2(x)+2sin(x)+1-4+4sin^2(x)=0$ $5sin^2(x)+2sin(x)-3=0$ $sin(x)=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(2) \pm \sqrt{(2)^2 - 4 (5)(-3)}}{2(5)}=\frac{3}{5},-1$ $sin(x)=\frac{3}{5}=0.6$ $x=sin^{-1}(\frac{3}{5})$ $x=2.4981$ $sin(x)=-1=\;\;\;\;$ $x=sin^{-1}(-1)$ $x=4.7124$ $x=\{2.4981,4.7124\}$
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