[tex]\sin^2x=2\sin x+3[/tex]
[tex]\sin^2x-2\sin x-3=0[/tex]
[tex](\sin x-3)(\sin x+1)=0[/tex]
which means either [tex]\sin x=3[/tex] or [tex]\sin x=-1[/tex]. The equation has no solution, since [tex]\sin x[/tex] is always bounded between -1 and 1. The second has one solution at [tex]x=-\dfrac\pi2[/tex], and any number of complete revolutions will also satisfy this equation, so in general the solution would be [tex]-\dfrac\pi2+2k\pi[/tex] where [tex]k[/tex] is any integer.
So you could choose [tex]A=-\dfrac\pi2[/tex] and [tex]B=2[/tex].
