Задание № 1 (2 балла) Решите уравнение: 1) 3cos²2x + 2sin 22x = 2,5 sin 4x; 2) 6sin²(x/2-π/6) + 0,5 sin(x-π/3) = 2+cos²(π/6-x/2); 3) cos x (2 cos x + tg x) = 1; 4) tg(2π-x) cos (3π/2+2x) = sin(-π/2).

Решение заданий №1: 1) 3cos²2x + 2sin²2x = 2,5 sin 4x 3cos²2x + 2sin²2x = 5 sin 2x cos 2x 3cos²2x + 2sin²2x - 5 sin 2x cos 2x = 0 3cos²2x + 2sin²2x - 5 \(2 sin x cos x\) cos 2x = 0 Разделим обе части уравнения на cos²2x (предполагая, что cos 2x ≠ 0): 3 + 2tg²2x - 5tg 2x = 0 2tg²2x - 5tg 2x + 3 = 0 Пусть t = tg 2x, тогда: 2t² - 5t + 3 = 0 D = (-5)² - 4 \(2\)(3) = 25 - 24 = 1 t₁ = \(\frac{5 + 1}{4}\) = 1,5 t₂ = \(\frac{5 - 1}{4}\) = 1 Тогда: tg 2x = 1,5 или tg 2x = 1 2x = arctg 1,5 + πn, n ∈ Z или 2x = \(\frac{π}{4}\) + πk, k ∈ Z x = \(\frac{1}{2}\) arctg 1,5 + \(\frac{πn}{2}\), n ∈ Z или x = \(\frac{π}{8}\) + \(\frac{πk}{2}\), k ∈ Z 2) 6sin²(x/2-π/6) + 0,5 sin(x-π/3) = 2+cos²(π/6-x/2); 6sin²(x/2-π/6) + 0,5 sin(x-π/3) = 2+cos²(-(x/2-π/6)); 6sin²(x/2-π/6) + 0,5 sin(x-π/3) = 2+cos²(x/2-π/6); Используем формулу sin(x - y) = sin x cos y - cos x sin y: sin(x - π/3) = sin x cos(π/3) - cos x sin(π/3) = \(\frac{1}{2}\) sin x - \(\frac{\(\sqrt{3}\)}{2}\) cos x sin²(x/2-π/6) = \(\frac{1 - cos(x - π/3)}{2}\) cos²(x/2-π/6) = \(\frac{1 + cos(x - π/3)}{2}\) Подставляем: 6 \(\frac{1 - cos(x - π/3)}{2}\) + 0,5 sin(x-π/3) = 2 + \(\frac{1 + cos(x - π/3)}{2}\) 3 - 3cos(x - π/3) + 0,5 sin(x-π/3) = 2 + 0,5 + 0,5 cos(x - π/3) -3,5 cos(x - π/3) + 0,5 sin(x-π/3) = -0,5 Умножим на 2: -7 cos(x - π/3) + sin(x-π/3) = -1 Пусть y = x - π/3: sin y - 7 cos y = -1 sin y = 7 cos y - 1 Используем основное тригонометрическое тождество sin²y + cos²y = 1: \((7 cos y - 1)²\) + cos²y = 1 49 cos²y - 14 cos y + 1 + cos²y = 1 50 cos²y - 14 cos y = 0 2 cos y (25 cos y - 7) = 0 cos y = 0 или 25 cos y - 7 = 0 cos y = 0 или cos y = \(\frac{7}{25}\) y = \(\frac{π}{2}\) + πn, n ∈ Z или y = ±arccos(\( \frac{7}{25}\)) + 2πk, k ∈ Z Тогда: x - π/3 = \(\frac{π}{2}\) + πn, n ∈ Z или x - π/3 = ±arccos(\( \frac{7}{25}\)) + 2πk, k ∈ Z x = \(\frac{5π}{6}\) + πn, n ∈ Z или x = \(\frac{π}{3}\) ± arccos(\( \frac{7}{25}\)) + 2πk, k ∈ Z 3) cos x (2 cos x + tg x) = 1 cos x (2 cos x + \(\frac{sin x}{cos x}\)) = 1 2 cos²x + sin x = 1 2 (1 - sin²x) + sin x = 1 2 - 2sin²x + sin x = 1 -2sin²x + sin x + 1 = 0 2sin²x - sin x - 1 = 0 Пусть t = sin x, тогда: 2t² - t - 1 = 0 D = (-1)² - 4 \(2\)(-1) = 1 + 8 = 9 t₁ = \(\frac{1 + 3}{4}\) = 1 t₂ = \(\frac{1 - 3}{4}\) = -0,5 Тогда: sin x = 1 или sin x = -0,5 x = \(\frac{π}{2}\) + 2πn, n ∈ Z или x = -\(\frac{π}{6}\) + 2πk, k ∈ Z, x = \(\frac{7π}{6}\) + 2πm, m ∈ Z 4) tg(2π-x) cos (3π/2+2x) = sin(-π/2) -tg(x) sin(2x) = -1 \(\frac{-sin(x)}{cos(x)}\) \(2sin(x)cos(x)\) = -1 -2sin²(x) = -1 sin²(x) = \(\frac{1}{2}\) sin x = ± \(\frac{1}{\(\sqrt{2}\)}\) x = ± \(\frac{π}{4}\) + πn, n ∈ Z