2cos^2x=1+4sin4x ПОМОГИТЕ

2cos²x=1+4sin4x2(1+cos2x/2)=1+8sin2xcos2x1+cos2x=1+8sin2xcos2x1+cos2x-1+8sin2xcos2x=0cos2x+8sin2xcos2x=0cos2x(1+8sin2x)=0cos2x=02x= π/2+πk |:2x= π/4+πk/2sin2x=-1/82x=(-1)^k+1 arcsin1/8+πn |:2x=(-1)^k+1 arcsin1/16+πn/2