SOLUTION: solve :(sin B)(sinB)-3sin B+2=0 where 0 < B < 2pie .

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Question 410944: solve :(sin B)(sinB)-3sin B+2=0 where 0 < B < 2pie .
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
Let z+=+sin%28B%29. Then, we have a quadratic

z%5E2+-+3z+%2B+2+=+0. Factoring, this is equal to

%28z-2%29%28z-1%29+=+0 --> z = 2 or z = 1, sin(B) = 2 or sin(B) = 1 by back-substituting. However, sin(B) cannot be 2 since it is bounded by -1+%3C=+sin%28B%29+%3C=+1. Therefore, the only solutions occur when sin%28B%29+=+1 --> B+=+pi%2F2.