General solution of the equation sin4x+cos4x=sinxcosx is

We have sin4x+cos4x=sinxcosx⇒1−2sin2xcos2x=sinxcosx⇒1−2t2=t where t=sinxcosx⇒2t2+t−1=0⇒2t2+2t−t−1=0⇒2t(t+1)−1(t+1)=0⇒(2t−1)(t+1)=0⇒t=12or  t=−1⇒sinxcosx=12   or   sinxcosx=−1⇒sin2x=1    or    sin2x=−2not possible⇒2x=2nπ+π2⇒x=nπ+π4,n∈z                  =(4n+1)π4