SOLUTION: sin (2x) = cos (2x) between (0, 2pi)

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Question 551337: sin (2x) = cos (2x) between (0, 2pi)
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
sin (2x) = cos (2x) between (0, 2pi)
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sin^2 = cos^2 = 1 - sin^2
2sin^2 = 1
sin^2(2x) = 1/2
sin%282x%29+=+sqrt%282%29%2F2
sin%282x%29+=+-sqrt%282%29%2F2
2x = pi/4, 3pi/4, 5pi/4, 7pi/4, 9pi/4, 11pi/4, 13pi/4, 15pi/4
x = pi/8, 3pi/8, 5pi/8, 7pi/8, 9pi/8, 11pi/8, 13pi/8, 15pi,8
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x = pi/8 + n*pi/4, n = 0 to 7