sin3xcosx+sin2xcos2x+sinxcos3x=1
⇒sinxcosx(sin2x+sinxcosx+cos2x)=1
⇒2sin2x(1+2sin2x)=1
⇒sin2x(2+sin2x)=4
⇒sin22x+2sin2x−4=0
⇒sin2x=2−2±4+16
⇒sin2x=−1±5
This is not possible. \left(Since - 1 \le \sin 2x \le 1\right)</p><p>Hencethegivenequationhasnosolution.</p>