2cos^2x=1+4sin4x ПОМОГИТЕ

2cos²x=1+4sin4x
2(1+cos2x/2)=1+8sin2xcos2x
1+cos2x=1+8sin2xcos2x
1+cos2x-1+8sin2xcos2x=0
cos2x+8sin2xcos2x=0
cos2x(1+8sin2x)=0
cos2x=0
2x= π/2+πk |:2
x= π/4+πk/2
sin2x=-1/8
2x=(-1)^k+1 arcsin1/8+πn |:2
x=(-1)^k+1 arcsin1/16+πn/2