SOLUTION: solve for x in [0,2pi] cosx-sinx=-1

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Question 678142: solve for x in [0,2pi] cosx-sinx=-1
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
solve for x in [0,2pi] cosx-sinx=-1
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cosx-sinx=-1
sqrt%281-sin%5E2%28x%29%29+=+sin%28x%29+-+1
1+-+sin%5E2%28x%29+=+sin%5E2%28x%29+-+2sin%28x%29+%2B+1
2sin%5E2%28x%29+-+2sin%28x%29+=+0
sin%5E2%28x%29+-+sin%28x%29+=+0
sin%28x%29%2A%28sin%28x%29+-+1%29+=+0
sin(x) = 0 --> x = 0, pi
sin(x) = 1 --> x = pi/2
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Check:
cos(0) - sin(0) = 1-0 = +1 (extraneous, reject)
cos(pi) - sin(pi) = -1 - 0 = -1
cos(pi/2) - sin(pi/2) = 0 -1 = -1
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x = pi/2, x = pi