Question

If $A$ and $B$ are acute angles such that $3{\sin }^{2}A+2{\sin }^{2}B=1$ and $3\sin 2A-2\sin 2B=0$, then $(A+2B)$ is equal to :

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{2}$
• D. none of these

Answer 1

The correct option is C $\frac{\pi }{2}$

$3{\sin }^{2}A+2{\sin }^{2}B=1$
$\Rightarrow 3{\sin }^{2}A=1-2{\sin }^{2}B$
$\Rightarrow 3{\sin }^{2}A=\cos 2B$ ....(1)
$3\sin 2A-2\sin 2B=0$
$\Rightarrow \sin 2A=\frac{2}{3}\sin 2B$ ....(2)
Now, $\cos (A+2B)=\cos A\cos 2B-\sin A\sin 2B$
$\Rightarrow \cos (A+2B)=3\cos A{\sin }^{2}A-\frac{3}{2}\sin A\sin 2A$ (using (1) and (2))
$\cos (A+2B)=3\cos A{\sin }^{2}A-3\cos A{\sin }^{2}A$
$\Rightarrow \cos (A+2B)=0$
$\Rightarrow A+2B=\frac{\pi }{2}or\frac{3\pi }{2}$
Since, $A$ and $B$ are acute angles
Hence, $A+2B=\frac{\pi }{2}$