Find all solutions of the equation sin⁡2x=2sin⁡x+3.\displaystyle {{\sin}^{{2}}{x}}={2}{\sin{{x}}}+{3}.
The answer is A+Bkπ\displaystyle {A}+{B}{k}\pi where k\displaystyle {k} is any integer and 0<A<2π\displaystyle {0}<{A}<{2}\pi,
A=\displaystyle {A}=   , B=\displaystyle {B}=   .

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