Let y=sinxcos3xsin3xcosx. Then

Given that , y=sinxcos3xsin3xcosx ⇒y(3sinx−4sin3x)cosx=sinx(4cos3x−3cosx)                                  (∵ sin3x=3sinx−4sin3x,  cos3x=4cos3x−3cosx) ⇒y(3−4sin2x)cosxsinx=sinxcosx(4cos2x−3) ⇒y(3−4sin2x)=4cos2x−3 [∵sinx≠0,cosx≠0] ⇒4(y−1)sin2x+(1−3y)=0        (∵ cos2x=1−sin2x) ⇒sin2x=3y−14(y−1)          ∵00 and 3y−14(y−1)−1<0 ⇒(3y−1)(y−1)>0  and (y−3)(y−1)>0 ⇒y<13 or y<1  and  y<1  or  y>3 ⇒y<13 or y>3