SOLUTION: solve: {{{2sin(2x)+cos(2x)+1=0}}} for {{{0<=x<=360}}}

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Question 884078: solve: 2sin%282x%29%2Bcos%282x%29%2B1=0 for 0%3C=x%3C=360
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
solve: 2sin%282x%29%2Bcos%282x%29%2B1=0 for 0%3C=x%3C=360
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2sin%282x%29%2Bcos%282x%29%2B1=0
cos%282x%29%2B1=-2sin%282x%29
cos%282x%29%2B1=-2sqrt%281-cos%5E2%282x%29%29
square both sides
cos^2(2x)+2cos(2x)+1=4(1-cos^2(2x)
cos^2(2x)+2cos(2x)+1=4-4cos^2(2x)
5cos^2(2x)+2cos(2x)-3=0
(5cos(2x)-3)(cos(2x)+1)=0
..
(5cos(2x)-3)=0
cos(2x)=3/5
2x≈53.13˚ and 306.87
x≈26.57˚ and 153.43˚
reject, (extraneous roots)
..
Solution:
cos(2x)+1=0
cos(2x)=-1
2x=180˚
x=90˚