Question

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

Answer 1

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