Помогите решить cos12x-sin4x=0

sin(π/2-12x) -sin4x=02sin(π/4-8x)cos(π/4-4x)=0sin(π/4-8x)=0π/4-8x=πn-8x=-π/4+πnx=π/32+πn/8cos(π/4-4x)=0π/4-4x=π/2+πn-4x=-π/4+π/2+πn-4x=π/4+πnx=-π/16+πn/4