The general solution of sinx−3sin2x+sin3x= cosx−3cos2x+cos3x is [IIT 1989]

sinx−3sin2x+sin3x=cosx−3cos2x+cos3x        ⇒2sin2xcosx−3sin2x−2cos2xcosx+3cos2x=0 ⇒ sin2x(2cosx−3)−cos2x(2cosx−3)=0 ⇒(sin2x−cos2x)(2cosx−3)=0⇒sin2x=cos2x ⇒ 2x=2nπ±(π2−2x) i.e.,  x=nπ2+π8.