Question

If A and B are acute positive angles satisfying the equations $3{\sin }^{2}A+2{\sin }^{2}B=1$ and $3\sin 2A-2\sin 2B=0$, then $A+2B$ is equal to

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. $3\frac{\pi }{4}$
• D. $\frac{2\pi }{3}$

Answer 1

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

$3{\sin }^{2}A+2{\sin }^{2}B=1\Rightarrow 3{\sin }^{2}A=1-2{\sin }^{2}B=\cos 2B$ ...(1)
$3\sin 2A-2\sin 2B=0\Rightarrow \sin 2B=\frac{3}{2}\sin 2A$ ...(2)
From (1) and (2), we get
$\cos (A+2B)=\cos A\cos 2B-\sin A\sin 2B=\cos A\left(3{\sin }^{2}A\right)-\sin A\left(\frac{3}{2}\sin 2A\right)=3{\sin }^{2}A\cos A-3{\sin }^{2}A\cos A=0\Rightarrow A+2B=\frac{\pi }{2}$