$A$ and $B$ are positive acute angles satisfying the equations $3{\cos }^{2}A+2{\cos }^{2}B=4\text{ and }\frac{3\sin A}{\sin B}=\frac{2\cos \overline{B}}{\cos A},$ then $A+2B$ is equal to

Solution:

We have,

$\begin{array}{l} \frac{3\sin A}{\sin B}=\frac{2\cos B}{\cos A}\\ \Rightarrow \frac{3\sin A}{ \cos A}=\frac{2\cos B\sin B}{{\cos }^{2}A}\\ \Rightarrow \tan A=\frac{\sin 2B}{3{\cos }^{2}A}\end{array}$

$\begin{array}{l}\Rightarrow \tan A=\frac{\sin 2B}{\cos 2B}\times \frac{\cos 2B}{3{\cos }^{2}A}\\ \Rightarrow \tan A=\tan 2B\times \frac{2{\cos }^{2}B-1}{3{\cos }^{2}A}\\ \Rightarrow \tan A=\tan 2B\times \frac{\left(4-3{\cos }^{2}A-1\right)}{3{\cos }^{2}A}\\ \Rightarrow \tan A=\tan 2B\times \frac{3-3{\cos }^{2}A}{3{\cos }^{2}A}\\ \Rightarrow \tan A=\tan 2B\times {\tan }^{2}A\end{array}$

$\begin{array}{l}\Rightarrow \tan A\tan 2B=1\\ \Rightarrow \cos A\cos 2B=\sin A\sin 2B\\ \Rightarrow \cos (A+2B)=0\Rightarrow A+2B=\frac{\pi }{2}\end{array}$