$A$ and $B$ are the positive acute angles satisfying the equations $3\cos^2 A + 2\cos^2 B=4$ $A$ and $B$ are the positive acute angles satisfying the equations $3\cos^2 A + 2\cos^2 B=4$ and $\dfrac {3\sin A}{\sin B}=\dfrac {2\cos B}{\cos A}$. Then $A+2B$ is equal to:
$1$. $\dfrac {\pi}{4}$
$2$. $\dfrac {\pi}{3}$
$3$. $\dfrac {\pi}{6}$
$4$. $\dfrac {\pi}{2}$.
My Attempt
$$\dfrac {3\sin A}{\sin B}=\dfrac {2\cos B}{\cos A}$$
$$3\sin A.\cos A= 2\cos B.\sin B$$
$$\dfrac {3}{2} \sin 2A=2\sin 2B$$.
How do I proceed further?

From the first equation
$$\cos  2B = 2\cos ^2B - 1=\left(4-3\cos ^2A\right)-1=3\sin ^2A$$
From the second equation
$$\sin  2B = \frac{3}{2}\sin  2A$$
Replace these values in 
$$\cos (A+2B)=\cos  A \cos  2B - \sin  A \sin  2B=$$
$$=\cos  A \left(3\sin ^2A\right)-\sin  A \left(\frac{3}{2}\sin  2A\right)=0$$
$$A+2B=\frac{\pi }{2}$$
