Question

If \( A \) and \( B \) are positive acute angles satisfying
\( 3\cos^2 A + 2\cos^2 B = 4 \) and \( \dfrac{3\sin A}{\sin B} = \dfrac{2\cos B}{\cos A} \),
then the value of \( A + 2B \) is equal to:

• A. \(\dfrac{\pi}{4}\)
• B. \(\dfrac{\pi}{2}\)
• C. \(\dfrac{\pi}{3}\)
• D. (π)/(6)

Hint:

Convert products of sine and cosine into sin 2θ.

Answer 1

The Correct Option is B

Step 1: From the second condition:
\( 3\sin A \cos A = 2\sin B \cos B \)
\( \Rightarrow 3\sin 2A = 2\sin 2B \) 

Step 2: Solving along with first equation gives:
\( A = \dfrac{\pi}{6}, \; B = \dfrac{\pi}{6} \) 

Step 3:
\( A + 2B = \dfrac{\pi}{6} + 2 \cdot \dfrac{\pi}{6} = \dfrac{\pi}{2} \)