Question 608825: Using the unit circle to solve for x in radians, and withthe restriction of [0,2pi), how would you solve sin2xsinx-cosx=0?
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Using the unit circle to solve for x in radians, and withthe restriction of [0,2pi), how would you solve sin2xsinx-cosx=0?
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sin(2x) = 2sin(x)cos(x)
sin2xsinx-cosx=0
2sin^2(x)cos(x) - cos(x) = 0
cos(x) = 0
x = pi/2, 3pi/2
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2sin^2(x) = 1
sin(x) = sqrt(2)/2
x = pi/4, 3pi/4
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