Question 809492: Solve 1+cos(2x)=2sin^2(2x) for x, where 0 ≤ to x < 2π
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Solve 1+cos(2x)=2sin^2(2x) for x, where 0 ≤ to x < 2π
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1+cos(2x)=2sin^2(2x)
1+cos(2x)=2(1-cos^2(2x)
1+cos(2x)=2-2cos^2(2x)
2cos^2(2x)+cos(2x)-1=0
(2cos(2x)-1)(cos(2x)+1)=0
..
2cos(2x)-1=0
cos(2x)=1/2
2x=π/3, 5π/3
x=π/6, 5π/6
..
cos(2x)+1=0
cos(2x)=-1
2x=π
x=π/2
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