# 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)

SAGAR SINGH - IIT DELHI
879 Points
13 years ago

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

All the best.

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Sagar Singh

B.Tech, IIT Delhi

kaushik B
13 Points
5 years ago
There is a little typo error in the solution. (square is showing as 2A)
It is given 3Sin2A+2Sin2B=1 => 3Sin2A = 1-2Sin2B => 3Sin2A = Cos 2B.....(i)
3Sin2A= 2Sin2B (given) => 3/2Sin2A=Sin2B…...(ii)
We need to find Cos(A+2B) =?   Cos(A+2B) = cosAcos2B – SinASin2B = cosA(3Sin2A)- sinA(3/2Sin2A)
= 3SinA*(1/2)*2SinAcosA – 3/2SinAsin2A = (3/2)SinAsin2A-(3/2)SinASin2A = 0= Sin(π/2) therefore
A+2B= π/2