sin^2 x + 2 sin (π - x) cos x - 3 cos^2 (2π - x) = 0

Sin^2 x + 2 sin (π - x) * cos x - 3 cos^2 (2π - x) = 0sin² x + 2 sin x * cos x - 3 cos² x = 0 Пояснение: sin (pi - x) = sin x   cos² (2pi - x) = cos² xsin² x + 2 sin x * cos x - 3 cos² x = 0   | : cos²x ≠ 0tg² x + 2 tg x - 3 = 0  Вводим замену  tg x = tРешаем квадратное уравнениеt² + 2t - 3 = 0D = b² - 4ac = 2² - (-4*1*3) = 4 + 12 = 16  √D = 4t1 = (-2+4)/2 = 1t2 = (-2-4)/2 = -3tg x = t1) tg x = 1 x = pi/4 + pik, k ∈ Z2) tg x = -3x = -arctg3 + pik, k ∈ ZОТВЕТ: pi/4 + pik, k ∈ Z; -arctg3 + pik, k ∈ Z