<s>[INST] 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}
 [/INST] Note that:
$$\begin{cases}\cos2B=2\cos^2B-1=3-3\cos^2A\\ \sin2B=2\sin B\cos B=3\sin A\cos A\end{cases} \Rightarrow \\
\cos^22B+\sin^22B=9-18\cos^2A+9\cos^4A+9\sin^2A\cos^2A \Rightarrow \\
1=9-18\cos^2A+9\cos^4A+9(1-\cos^2A)\cos^2A \Rightarrow \\
1=9(1-\cos^2A) \Rightarrow \\
\sin^2A=\frac19 \Rightarrow \\
\sin A=\frac13 \ \ (\text{because} \ 0<A<\frac{\pi}{2}).$$
You can plug this into your relation to get what you want:

 $\sin(A+2B)=3\sin A=3\cdot \frac13=1 \Rightarrow A+2B=\frac{\pi}{2}.$

</s>