The number of x∈[0,2π] for which |√2sin4x+18cos2x−√2cos4x+18sin2x|=1 is:

sin2x+cos2x=1
sin4x+cos4x+2sin2xcos2x=1
sin4x+cos4x=1−2sin2xcos2x…[1]
sin2x+cos2x=1
sin6x+cos6x+3sin2xcos2x(sin2x+cos2x)=1
sin6x+cos6x=1−3sin2xcos2x…[2]
|√2sin4x+18cos2x−√2cos4x+18sin2x|=1
Squaring above equation
∴2sin4x+18cos2x+2cos4x+18sin2x−2(√2sin4x+18cos2x)(√2cos4x+18sin2x)=1
∴2(sin4x+cos4x)+18(cos2x+sin2x)−2(√2sin4x+18cos2x)(√2cos4x+18sin2x)=1
∴2(1−2sin2xcos2x)+18−2(√2sin4x+18cos2x)(√2cos4x+18sin2x)=1 ( [Using[1] )
∴20−4sin2xcos2x−2(√2sin4x+18cos2x)(√2cos4x+18sin2x)=1
∴19−4sin2xcos2x=2(√2sin4x+18cos2x)(√2cos4x+18sin2x)
Squaring above equation
∴361+16sin4xcos4x−76sin2xcos2x=4(4sin4xcos4x+36cos6x+36sin6x+324sin2xcos2x)
∴361+16sin4xcos4x−76sin2xcos2x=16sin4xcos4x+144cos6x+144sin6x+1296sin2xcos2x
∴361=144(cos6x+sin6x)+1372sin2xcos2x
∴361=144(1−3sin2xcos2x)+1372sin2xcos2x ( Using [2] )
∴361=144+1016sin2xcos2x
∴217=1016sin2xcos2x
∴217=254sin22x
∴sin2x=±√217/254
Let sin⁻¹(√217/254)=θ
sin2x=±√217/254
∴2x=sin⁻¹(√217/254)
∴2x=θ,π−θ,2π+θ,3π−θ
∴x=θ/2,π−θ/2,2π+θ/2,3π−θ/2
2x=sin⁻¹(−√217/254)
∴2x=π+θ,2π−θ,3π+θ,4π−θ
∴x=π+θ/2,2π−θ/2,3π+θ/2,4π−θ/2
So, total number of solutions =8
So, the answer is option (D)