If `Aa n dB` are acute positive angles satisfying the equations `3sin^2A+2sin^2B=1` and `3sin2A-2sin2B=0,` then `A+2B` is equal to (a)`pi` (b) `pi/2` (c) `pi/4` (d) `pi/6`

Explanation

`3sin^2A+2sin^2B = 1`

`=>3sin^2A = 1-2sin^2B`

`=>3sin^2A = cos2B->(1)`

Now,

`3sin2A-2sin2B = 0`

`=>3sin2A = 2sin2B`

`=>sin2B = 3/2sin2A->(2)`

Now, `cos(A+2B) = cosAcos2B-sinAsin2B`

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