**Find the solutions of the equation in [0,2pi) 2sin^2(u)=1-sin(u)**

**Answer:**

If we rearrange it

2 Sin^2u + Sinu -1 = 0 , its a quadratic simply solve like a quadratic considering sinu as x

2sin^2u + 2sinu -sinu -1 = 0

2sinu (sinu +1) - 1(sinu+1) = 0

(sinu+1)*(2sinu-1) = 0

so solutions are

sinu +1 =0

sinu = -1

=> within the bracket [0,2pi) only one solution u = 270 degrees

2sinu -1 =0

=> sinu =1/2

within the bracket [0,2pi) two solutions u = 30 degrees and 150 degrees

so total solutions in [0,2pi) are three viz u = 30 degree, 150 degree , 270 degree