SOLUTION: Solve the following equations for (0,2pi) 7.sin^2x-2sinx+1=0

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Question 1071376: Solve the following equations for (0,2pi) 7.sin^2x-2sinx+1=0
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Is it 7sin^2(x) ?
Or problem #7 ?
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sin^2x-2sinx+1=0
(sin(x) - 1)^2 = 0
sin(x) = 1
x = pi/2