Решите уравнение 6cos^2 2x+22sin^2x-7=0

6cos²2x + 22sin²x - 7 = 06cos²2x - 11•(1 - 2sin²x) + 4 = 0

cos2x = 1 - 2sin²x

6cos²2x - 11cos2x + 4 = 0Пусть cos2x = a, a ∈ [ - 1 ; 1 ], тогда6а² - 11а + 4 = 0D = (-11)² - 4•6•4 = 121 - 96 = 25 = 5²a₁ = (11 - 5)/12 = 6/12 = 1/2a₂ = (11 + 5)/12 = 16/12 = 4/3 > 1 - не подходитcos2x = 1/22x = ± π/3 + 2πnx = ± π/6 + πn, n ∈ ZОТВЕТ: ± π/6 + πn, n ∈ Z