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

02pi-2sin2u1-sinu

#1

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