If A and B are two positive angles such that 3cos²A+2cos²B = 4 and 3sinA/sinB = 2cosB/cos A, then A+2B = ?

3cos²A + 2cos²B = 4           ................1

cos2A = 2cos²A - 1

cos2B = 2cos²B - 1

eq 1 becomes

3(1+cos2A)/2 + (1+cos2B) = 4                    .................2

3sinA/sinB = 2cosB/cosA

3sinAcosA =2sinB cosB                         (sin2@ = 2sin@cos@)

3sin2A = 2sin2B               .......................3

squaring eq 3

9sin²2A = 4sin²2B

9(1-cos²2A) = 4(1-cos²2B)

 9cos²2A - 4cos²2B = 5              ....................4

after solving 4 & 2 we get

cos2A = 7/9 & cos2B = 1/3

cos2A = 2cos²A-1 so

cosA = 2root2/3 , cos2B = 1/3                                  

A + 2B = cos^-12root2/3 + cos^-11/3                              (cos^-1x+cos^-1y = cos^-1{xy-[(1-x²)(1-y²)]^1/2}

         =cos^-10 = pi/2 = 90^o ...