Sin6x+cos7x=0.С пояснениями пожалуйста

sin6x+sin(π/2-7x)=02sin(13x/2-π/4)cos(π/4-x/2)=0sin(13x/2-π/4)=013x/2-π/4=πn,n∈Z13x/2=π/4+πn,n∈Zx=π/26+2πn/13,n∈Zcos(π/4-x/2)=sin(x/2+π/4)=0x/2+π/4=πk,k∈Zx/2=-π/4+πk,k∈Zx=-π/2+2πk,k∈Z