trigonometry

The question which you have posted is wrong. The correct statement of the question is as follows:

3sin²A+ 2sin²B =1     ... (1)

and 3 sin2A -2sin2B =0  ... (2)

 

Following is the solution:

 

From equation (1), we have:

3sin²A = 1 - 2sin²B = cos 2B    ... (3)

From (2), we have:

3 sin2A = 2 sin 2B                   ... (4)

 

Consider cos (A + 2B) = cosA cos 2B - sin A sin 2B

                               = cosA (3sin²A) - sin A (3 sin 2A/2)      [Using (3) and (4)]

                               = 3 cosA sin²A - sin A (3 sin A cosA)     [Since, sin 2A = 2 sinA cos A]

                               = 3 cosA sin²A - 3 sin² A cosA

                               = 0

Thus, A + 2B = ?/ 2       

[Since, cos ?/ 2 = 0]