Find $A+2B$ if $3\cos^2A+2\cos^2B=4$ and $\frac{3\sin A}{\sin B}=\frac{2\cos B}{\cos A}$ 
$A$ and $B$ are positive acute angles satisfying $3\cos^2A+2\cos^2B=4$ and $\dfrac{3\sin A}{\sin B}=\dfrac{2\cos B}{\cos A}$, then find the value of $A+2B$ ?

My Attempt
$\cos2B=2\cos^2B-1=3-3\cos^2A$ and $\sin2B=2\sin B\cos B=3\sin A\cos A$
\begin{align}
\sin(A+2B)&=\sin A\cos2B+\cos A\sin2B\\
&=\sin A.[3-3\cos^2A]+\cos A.[3\sin A\cos A]=3\sin A\\
\sin(A+2B)=3\sin A\implies \color{red}{?}
\end{align}
How do I prove that $A+2B=\dfrac{\pi}{2}$ from $\sin(A+2B)=3\sin A$ ?
What I know
\begin{align}
\cos(A+2B)&=\cos A\cos2B-\sin A\sin2B\\
&=\cos A.[3-3\cos^2A]+\sin A.[3\sin A\cos A]\\
&=3\cos A-3\cos^3A+3\sin^2A\cos A\\
&=3\cos A-3\cos A.[\cos^2A+\sin^2A]=0\implies \boxed{A+2B=\dfrac{\pi}{2}}
\end{align}

Hint:
Using double angle formula, 
$$2\sin2B=6\sin A\cos A=3\sin2A$$
$$2\cos2B=6(1-\cos^2A)=6(2\sin^2A)=3(1-\cos2A)$$
Squaring and adding we get $\dfrac49=2-2\cos2A=2(2\sin^2A)$
$\implies\sin A=\dfrac13,\cos A=+\dfrac{2\sqrt2}3$ as $0<A<\dfrac\pi2$
$$\sin2B=3\sin A\cos A=\dfrac{2\sqrt2}3,$$
$$\cos2B=3\sin^2A=\dfrac13$$
Hope you can take it from here using $\cos(2B+A)$ or $\sin(2B+A)$!
