1) Решите уравнение 3 sin²x = cos²x. 2) Найдите корни этого уравнения, принадлежащие отрезку [3π/2; 3π].

1. Преобразуем уравнение: 3sin²x = cos²x => 3tan²x = 1 => tan²x = 1/3 => tanx = ±1/√3.
2. Общее решение: x = π/6 + πn и x = -π/6 + πn, где n ∈ Z.
3. Корни на отрезке [3π/2; 3π]: 3π/2 ≤ π/6 + πn ≤ 3π => 3π/2 - π/6 ≤ πn ≤ 3π - π/6 => 8π/3 ≤ πn ≤ 17π/6 => 8/3 ≤ n ≤ 17/6 => n=2. x = π/6 + 2π = 13π/6.
3π/2 ≤ -π/6 + πn ≤ 3π => 3π/2 + π/6 ≤ πn ≤ 3π + π/6 => 10π/3 ≤ πn ≤ 19π/6 => 10/3 ≤ n ≤ 19/6 => n=2. x = -π/6 + 2π = 11π/6.
Проверим второй интервал: 3π/2 ≤ π/6 + πn ≤ 3π => n=2, x=13π/6.
3π/2 ≤ -π/6 + πn ≤ 3π => n=3, x = -π/6 + 3π = 17π/6.
Ответ: 13π/6, 17π/6.