Решите, пожалуйста, срочно. cos4x-cosx*cos3x=0

cosx*cos3x=0.5(cos2x+cos4x) cos4x-cos2x/2-cos4x/2=1/2cos4x-cos2x/2=1/2(2cos²2x-1)-1/2 cos2x cos²2x-1/2cos2x-1/2=0 cos2x=z 3z²-0.5z-1/2=0 6z²-z-1=0 z=1/2    z=-1/3 cos2x=1/2   2x=+-π/3+2πn   x=+-π/6+πn cos2x=-1/3  2x=+-arccos(5/6π)+2πn   x=+-1/2*arccos5π/6+πn n∈Z