2sin^2(u)=−1−3sin(u)

Question

Find the solutions of the equation that are in the interval [0, 2π).

Answer 1

2sin²(u)+3sin(u)+1=0=(2sin(u)+1)(sin(u)+1),

so sin(u)=-½ or sin(u)=-1

sine is negative in the third and fourth quadrants

sin(π+π/6)=-sin(π/6)=-½=sin(2π-π/6), so u=7π/6 or 11π/6.

sin(3π/2)=-1, so a third solution is u=3π/2.