$A$ 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?
 A: 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 
$$\cos (A+2B)=\cos  A \cos  2B - \sin  A \sin  2B=$$
$$=\cos  A \left(3\sin ^2A\right)-\sin  A \left(\frac{3}{2}\sin  2A\right)=0$$
$$A+2B=\frac{\pi }{2}$$
A: HINT: we have:
$$\frac{3\sin\left(\text{A}\right)}{\sin\left(\text{B}\right)}=\frac{2\cos\left(\text{B}\right)}{\cos\left(\text{A}\right)}\space\Longleftrightarrow\space3\sin\left(\text{A}\right)\cos\left(\text{A}\right)=2\cos\left(\text{B}\right)\sin\left(\text{B}\right)\space\Longleftrightarrow\space$$
$$\frac{3\sin\left(2\text{A}\right)}{2}=\sin\left(2\text{B}\right)\tag1$$
Now, we also have:
$$3\cos^2\left(\text{A}\right)+2\cos^2\left(\text{B}\right)=1+3\cdot\frac{1+\cos\left(2\text{A}\right)}{2}+\cos\left(2\text{B}\right)=4\tag2$$
A: 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 $A+2B=\frac{\pi}{2}$
A: 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?
