The number of distinct solutions of the equation 54cos22x+cos4x+sin4x+cos6x+sin6x=2 in the interval [0,2π] is _________

54cos22x+(cos2x)²+(sin2x)²+(cos2x)³+(sin2x)³=2
54cos22x+(cos2x+sin2x)²-sin2x⋅cos2x+(cos2x+sin2x)(sin4x+cos4x-sin2x⋅cos2x)=2
→54cos22x+1-sin2x.cos2x+(sin4x+cos4x-sin2x⋅cos2x)=2
→54cos22x+1-sin2x.cos2x+(1-sin2x⋅cos2x)=2
→54cos22x-sin2x.cos2x=0
54-sin22x-sin2x⋅cos2x=0
→54-sin22x-sin2x⋅cos2x=0
→54sin22x=0
sin22x=1/2
→sin2x=±1/√2
∴x=π/8,3π/8,5π/8,7π/8,9π/8,11π/8,13π/8,15π/8
Hence number of solutions in the given interval is 8.