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