Verified Answer :

LammettHash

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

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 \(B=2\).

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