Find $A+2B$ if $3\cos^2A+2\cos^2B=4$ and $\frac{3\sin A}{\sin B}=\frac{2\cos B}{\cos A}$

Question

$A$ 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}

Answer 1

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:

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

Answer 2

Hint:

Using double angle formula,

$$2\sin2B=6\sin A\cos A=3\sin2A$$

$$2\cos2B=6(1-\cos^2A)=6(2\sin^2A)=3(1-\cos2A)$$

Squaring and adding we get $\dfrac49=2-2\cos2A=2(2\sin^2A)$

$\implies\sin A=\dfrac13,\cos A=+\dfrac{2\sqrt2}3$ as $0<A<\dfrac\pi2$

$$\sin2B=3\sin A\cos A=\dfrac{2\sqrt2}3,$$

$$\cos2B=3\sin^2A=\dfrac13$$

Hope you can take it from here using $\cos(2B+A)$ or $\sin(2B+A)$!