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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)
```
7 years ago

SAGAR SINGH - IIT DELHI
879 Points
```										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.
Win exciting gifts by answering the questions on Discussion Forum. So help discuss any                             query on askiitians forum and become an Elite     Expert        League            askiitian.

Sagar Singh
B.Tech, IIT Delhi

```
7 years ago
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