$A$ 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 

$$\cos (A+2B)=\cos  A \cos  2B - \sin  A \sin  2B=$$

$$=\cos  A \left(3\sin ^2A\right)-\sin  A \left(\frac{3}{2}\sin  2A\right)=0$$

$$A+2B=\frac{\pi }{2}$$
Hint: Square both sides of the second equation, and and replace $\sin^2 $ by $1 - \cos^2$, and coupled with the first solve a system of equations for $\cos A, \cos B$. Can you manager to proceed?