# Sines again

Geometry Level pending

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

$$A$$ and $$B$$ are the acute positive angles satisfying the equations above.

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

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