If 3 sin² A + 2 sin² B = 1 and 3 sin 2A − 2 sin 2B = 0, where A and B are acute angles then find the value of (A +2B)

Dear Nirabhra, 

sin2B=3/2sin2A …(i)

And from the first equation,

3sin2A=1-2sin2B=cos2B …(ii)

cos (A + 2B) = cos A. cos 2B – sin A.sin 2B

3cosA.sin2A-32sinA.sin2A

3cosA.sin2A-3sin2AcosA =0

cos (A+2B)=0

∴ A+2B=π2,3π2 …(iii)

Given that < A < π/2 and 0 < B < π/2

⇒ 0<A+2B<π+π2

⇒ 0<A+2B<3π2 …(iv)

∴ From (iii) and (iv),

A+2B=π2

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All the best.

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