The total number of solutions of sin4⁡x+cos4⁡x=sin⁡xcos⁡x in 0,2π is equal to

sin4⁡x+cos4⁡x=sin⁡xcos⁡x or  sin2⁡x+cos2⁡x2−2sin2⁡xcos2⁡x=sin⁡xcos⁡x or  1−sin2⁡2x2=sin⁡2x2 or  sin2⁡2x+sin⁡2x−2=0 or  (sin⁡2x+2)(sin⁡2x−1)=0 or  sin⁡2x=1 or  2x=(4n+1)π2,n∈Z or  x=(4n+1)π4,n∈Z =π4,5π4 (∵x∈[0,2π])Thus, there are two solutions