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

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