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

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`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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