$A$ 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?

The first condition gives $3\cos2A+2\cos2B=3$.
The second condition gives $3\sin2A=2\sin2B$ or 
$$9\sin^22A=4\sin^22B$$ or
$$9(1-\cos^22A)=4(1-\cos^22B)$$ or
$$9\cos^22A-4\cos^22B=5$$ or
$$(3\cos2A+2\cos2B)(3\cos2A-2\cos2B)=5$$ or
$$3(3\cos2A-2\cos2B)=5$$ or
$$3\cos2A-2\cos2B=\frac{5}{3},$$
which after summing with $$3\cos2A+2\cos2B=3$$ gives $6\cos2A=\frac{14}{3}$,
which says $\cos2A=\frac{7}{9}$ and $\cos2B=\frac{1}{3}$.
Thus,
$$\sin(A+2B)=\sin{A}\cos2B+\cos{A}\sin2B=$$
$$=\sqrt{\frac{1-\frac{7}{9}}{2}}\cdot\frac{1}{3}+\sqrt{\frac{1+\frac{7}{9}}{2}}\cdot\sqrt{1-\frac{1}{9}}=1,$$
which gives $A+2B=\frac{\pi}{2}$
