Задание № 3 (1 балл) Решите уравнение: \(\frac{1}{tg²x}\) - \(\frac{1}{sin x}\) - 1 = 0.

Решение задания №3: \(\frac{1}{tg²x}\) - \(\frac{1}{sin x}\) - 1 = 0 ctg²x - \(\frac{1}{sin x}\) - 1 = 0 \(\frac{cos²x}{sin²x}\) - \(\frac{1}{sin x}\) - 1 = 0 \(\frac{1 - sin²x}{sin²x}\) - \(\frac{1}{sin x}\) - 1 = 0 \(\frac{1 - sin²x - sin x - sin²x}{sin²x}\) = 0 \(\frac{-2sin²x - sin x + 1}{sin²x}\) = 0 -2sin²x - sin x + 1 = 0, sin x ≠ 0 2sin²x + sin x - 1 = 0 Пусть t = sin x, тогда: 2t² + t - 1 = 0 D = 1² - 4 \(2\)(-1) = 1 + 8 = 9 t₁ = \(\frac{-1 + 3}{4}\) = \(\frac{1}{2}\) t₂ = \(\frac{-1 - 3}{4}\) = -1 Тогда: sin x = \(\frac{1}{2}\) или sin x = -1 x = \(\frac{π}{6}\) + 2πn, n ∈ Z, x = \(\frac{5π}{6}\) + 2πk, k ∈ Z или x = -\(\frac{π}{2}\) + 2πm, m ∈ Z