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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`

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`3sin^2A+2sin^2B = 1`
`=>3sin^2A = 1-2sin^2B`
`=>3sin^2A = cos2B->(1)`
`3sin2A-2sin2B = 0`
`=>3sin2A = 2sin2B`
`=>sin2B = 3/2sin2A->(2)`
Now, `cos(A+2B) = cosAcos2B-sinAsin2B`
`= 0`
`:. cos(A+2B) = 0`
`=>A+2B = pi/2`
so, option `(b)` is the correct option.

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