Question

Find all solutions of the equation sin^2 x=2sinx+3

Answer 1

$\sin^2x=2\sin x+3$
$\sin^2x-2\sin x-3=0$
$(\sin x-3)(\sin x+1)=0$

which means either $\sin x=3$ or $\sin x=-1$. The equation has no solution, since $\sin x$ is always bounded between -1 and 1. The second has one solution at $x=-\dfrac\pi2$, and any number of complete revolutions will also satisfy this equation, so in general the solution would be $-\dfrac\pi2+2k\pi$ where $k$ is any integer.

So you could choose $A=-\dfrac\pi2$ and $B=2$.