SOLUTION: 2sin^2(u)=−1−3sin(u) Find the solutions of the equation that are in the interval [0, 2π). Please help!

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Question 743225: 2sin^2(u)=−1−3sin(u)

Find the solutions of the equation that are in the interval [0, 2π).
Please help!
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
2sin^2(u)=−1−3sin(u)
Find the solutions of the equation that are in the interval [0, 2π).
2sin%5E2%28u%29=-1-3sin%28u%29
2sin%5E2%28u%29%2B3sin%28u%29%2B1=0
(2sin(u)+1)(sin(u)+1)=0
..
2sin(u)+1=0
sin(u)=-1/2
u=7π/6, 11π/6
..
sin(u)+1=0
sin(u)=-1
u=3π/2