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If A and B be acute positive angles satisfying 3sin2⁡A+2sin2⁡B=1,3sin⁡2A−2sin⁡2B=0 then

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a

B=π4−A2

b

A=π4−2B

c

B=π2−A4

d

A=π4−B2

answer is A.

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Detailed Solution

sin⁡2B=(3/2)sin⁡2A,cos⁡2B=3sin2⁡A⇒tan⁡2B=cot⁡A⇒tan⁡Atan⁡2B=1⇒A+2B=π/2⇒B=π/4−A/2

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