SOLUTION: Find the EXACT solutions of the equation sin(x)+2cos^2(x)=1 that are in the interval [0, 2π).
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Question 524726: Find the EXACT solutions of the equation sin(x)+2cos^2(x)=1 that are in the interval [0, 2π).
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Find the EXACT solutions of the equation sin(x)+2cos^2(x)=1 that are in the interval [0, 2π)
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sin(x)+2cos^2(x)=1
sin(x)+2(1-sin^2(x)=1
sin(x)+2-2sin^2(x)=1
2sin^2(x)-sin(x)-1=0
(2sin(x)+1)(sin(x)-1)=0
(2sin(x)+1)=0
2sin(x)=-1
sin(x)=-1/2
x=7π/6 and 11π/6 (in quadrants III and IV where sin<0)
or
sin(x)-1=0
sin(x)=1
x=π/2
ans:
Solutions: 7π/6, 11π/6, and π/2
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