SOLUTION: Solve 1+cos(2x) = 2sin^2(2x) for x, where 0<= x < 2pie

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Question 850670: Solve 1+cos(2x) = 2sin^2(2x) for x, where 0<= x < 2pie

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Solve 1+cos(2x) = 2sin^2(2x) for x, where 0<= x < 2pie
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for: 0≤ 2x ≤ 4π
1%2Bcos%282x%29+=+2sin%5E2%282x%29
1%2Bcos%282x%29+=+2%281-cos%5E2%282x%29%29
1%2Bcos%282x%29+=+2-2cos%5E2%282x%29
2cos^2(2x)+cos(2x)-1=0
(2cos(2x)-1)(cos(2x)+1)=0
..
cos(2x)=1/2
2x=π/3, 5π/3, 7π/3, 11π/3
x=π/6, 5π/6, 7π/6, 11π/6
or
cos(2x)=-1
2x=π, 5π/2
x=π/2, 5π/4