SOLUTION: 2sin^2xcosx - cosx = 0 solve on the interval [0,2pi]

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Question 821209: 2sin^2xcosx - cosx = 0 solve on the interval [0,2pi]
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
2sin^2xcosx - cosx = 0 solve on the interval [0,2pi]
cosx(2sin^2x-1)=0
..
cosx=0
x=π/2, 3π/2
or
2sin^2x-1=0
sin^2x=1/2
sinx=±√(1/2)=±√2/2
x=π/4, 3π/4, 5π/4, 7π/4