Question

3sin²A - 2sinA - 5. ​

Answer 1

Answer:

2π

3sin2A+2sin2B=1

3sin2A=1−2sin2B

3sin2A=cos2B

3sinA=sinAcos2B     ....(1)

Also given 3sin2A=2sin2B

3(2sinAcosA)=2sin2B

3sinA=cosAsin2B      ....(2)

From (1) and (2), we get

cosAsin2B=sinAcos2B

⇒cos(A+2B)=0

⇒A+2B=2π