Question 743225
2sin^2(u)=−1−3sin(u) 
Find the solutions of the equation that are in the interval [0, 2π). 
{{{2sin^2(u)=-1-3sin(u)}}}
{{{2sin^2(u)+3sin(u)+1=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