sin x + √3 cos x = 0 /cosx tgx+√3=0 tgx=-√3 x=-π/3+πk,k∈z π/2<-π/3+πk<π *6/π 3<-1+6k<6 4/6<k<7/6 k=1 x=-π/3+π=2π/3=120гр
sin x + √3 cos x = 0 /cosx
tgx+√3=0
tgx=-√3
x=-π/3+πk,k∈z
π/2<-π/3+πk<π *6/π
3<-1+6k<6
4/6<k<7/6
k=1 x=-π/3+π=2π/3=120гр