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) (Show Source): You can put this solution on YOUR website!
Let . Then, we have a quadratic
. Factoring, this is equal to
--> 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 . Therefore, the only solutions occur when --> .
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