\(\begin{cases} 3\sin^2A + 2\sin^2B = 1 \\ 3\sin(2A)-2\sin(2B) =0 \end{cases} \).

\(A\) and \(B\) are the acute positive angles satisfying the equations above.

Then \(A+2B=\dfrac{\pi}{p}\). Find \(p\).

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