Question

Condition (0°≤A≤360°)
sin5A+sin3A=cos2A-cos6A​

Answer 1

sin5A+sin3A=cos2A-cos6A

2SIN((5A+3A)/2)COS ((5A-3A)/2)=2SIN((2A+6A)/2)SIN((2A-6A)/2)

`(2 \sin(4a )cosa)/( - 2 \sin(4a) \sin(2a) ) = 1`

`- ( \cos(a) )/(2 \sin(a) \cos(a ) ) = 1`

-1/2sinA=1

-1/2=sinA

sinA=sin(180+30),sin(330)

:.A=210,330