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Question

Let 0 < A,B < π2 satisfying the equation 3sin2A+2sin2B=1 and 3sin2A - 2sin2B = 0 then A + 2B is equal to


A

π

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B

π/2

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C

π/4

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D

2π

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Solution

The correct option is B

π/2


3sin2A=12sin2B=cos2B

Now cos(A+ 2B) = cosA cos2B - sinA sin2B

= 3cosA.sin2A32.sinAsin2A

= 3cosA.sin2A3sin2A.cosA=0

A + 2B = π2or3π2

Given that 0 < A < π2 and 0 < B < π2

0 < A + 2B < π+π2 Hence A + 2B = π2


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