SOLUTION: Solve for 0≤x<2pi: cosx·sinx = sin^2(x)

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Question 1138385: Solve for 0≤x<2pi: cosx·sinx = sin^2(x)
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
cos%28x%29%2Asin%28x%29+=+sin%5E2%28x%29
cos%28x%29%2Asin%28x%29+-+sin%5E2%28x%29=0
cos%28x%29%2Asin%28x%29+-+sin%28x%29%2Asin%28x%29=0
%28cos%28x%29-sin%28x%29%29%2Asin%28x%29=0
One of those 2 factor must be zero for the product to be zero.
cos%28x%29-sin%28x%29=0 if and only if cos%28x%29=sin%28x%29
When 0%3C=x%3C2pi ,
sin%28x%29=0 if and only if x=0 or x=pi , and
cos%28x%29=sin%28x%29 if and only if x=pi%2F4 or x=5pi%2F4