\sin\ \frac{\Pi}{6}\. \cos4x\ - \cos\\frac{\Pi}{6\}.\sin4x\ + \sin3x\ + \sinx\ =\frac{1}{2}\\frac{1}{2}\. \cos4x\ - \frac{\sqrt{3} }\{2}\. \sin4x\ +2\sin2x\.\cosx\ =\frac{1}{2}\(1- 2\sin^{2} 2x )\ + \sqrt{3}\ \sin2x\ .\cos2x\ + 2\sin2x\ . \cosx\ =\frac{1}{2}\-\sin ^{2} 2x\ + \sqrt{3}\\sin2x\ .\cos2x\ + 2\sin2x \. \cosx\ =0\\sin2x\(-\sin2x\ + \sqrt{3}\.\cos2x \+ 2\cosx)\ =\0 \