Найдите корни уравнения sin(4x/3+pi/6)=-1/2 , принадлежащие промежутку [-2pi; 2pi)

1)4x/3+π/6=7π/6+2n,n∈z 4x/3=π+2πn,n∈z x=3π/4+3πn/2,n∈z -2π≤3π/4+3πn/2<2π -8≤3+6n<8 -11≤6n<5 -11/6≤n<5/6 n=-1⇒x=3π/4-3π/2=-3π/4 n=0⇒x=3π/4 2)4x/3+π/6=11π/6+2πk,k∈z 4x/3=5π/3+2πk,k∈z x=5π/4+3πk/2;k∈z -2π≤5π/4+3πk/2<2π -8≤5+6k<8 -13≤6k<3 -13/6≤k<1/2 k=-2⇒x=5π/4-3π=-7π/4 k=-1⇒x=5π/4-3π/2=-π/4 k=0⇒x=5π/4