If A and B are acute positive angles satisfying the equations 3 sin2A+2 sin2B = 1 and 3 sin 2A -2 sin 2B=0, then A + 2B is equal to

3 sin2A + 2 sin2B = 1or 3sin2A=cos2BAlso 3 sin2A-2sin2B=0or sin⁡2B=32sin⁡2ANow, cos(A + 2B)=cosA cos2B - sinA sin2B=cos⁡A3sin2⁡A−sin⁡A32sin⁡2A=3sin2⁡Acos⁡A−3sin2⁡Acos⁡A=0∴A+2B=π2