Найдите корни уравнения 2 sin x + корень из 3 =0 принадлежащие отрезку[0, 2пи]

2sinx + √3 = 0
sinx = -√3/2
x = -π/3 + 2πk
x = -2π/3 + 2πk
1) 0 ≤ -π/3 + 2πk ≤ 2π
π/3 ≤ 2πk ≤ 7π/3
1/6 ≤ k ≤ 7/6
k = 1, x = -π/3 + 2π = 5π/3
2) 0 ≤ -2π/3 + 2πk ≤ 2π
2π/3 ≤ 2πk ≤ 8π/3
1/3 ≤ k ≤ 4/3
k = 1, x = -2π/3 + 2π = 4π/3

Ответ: 5π/3; 4π/3