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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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