**Correct option (C) π/2**

From the second equation, we have

sin2B = 3/2sin2A

and from the first equality

3sin^{2} A = 1 –2 sin2B = cos2B

Now cos (A + 2B) = cosA⋅cos2B – sinA⋅sin2B

= 3 cosA⋅sin^{2}A - 3/2 ⋅sinA⋅sin2A

= 3cosA⋅sin2A – 3sin2A⋅cosA = 0

⇒ A + 2B = π/2 or 3π/2

Given that 0 < A π/2 and 0 < B < π/2

⇒ 0 < A + 2B < π + π/2

Hence, A + 2B = π/2