SOLUTION: Solve the equations of [0, 2pi]: cosx(2sinx-1)=0 and (cos^2)x=(sin^2)x

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Question 870759: Solve the equations of [0, 2pi]:
cosx(2sinx-1)=0
and
(cos^2)x=(sin^2)x
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Solve the equations of [0, 2pi]:
cosx(2sinx-1)=0
cosx=0
x=π/2, 3π/2
2sinx-1=0
sinx=1/2
x=π/6, 3π/6
..
and
(cos^2)x=(sin^2)x
(cos^2)x-(sin^2)x=0
cos2x=0
2x=π/2, 3π/2
x=π/4, 3π/4