(a) We have,sin x = sin y ⇒ sin x - sin y = 02cos(2x+y)−sin(2x−y)=0⇒cos(2x+y)=0 or sin(2x−y)=0⇒2x+y=2(2n+1)π or 2x−y=nπ⇒x+y=(2n+1)π or x−y=2nπ⇒x=[(2n+1)π−y] or x=(2nπ+y)⇒ x = [(an odd multiple of π) - y] orx = [(an even multiple ofπ+ y]⇒ θ = nπ+ (-1)yn, where n ∈ Z