Решите уравнение: 6sin^2x + 7 cos x = 7

6sin²x + 7cosx = 7

6 (1 - cos²x) + 7cosx = 7

6 - 6cos²x + 7cosx - 7 = 0

-6cos²x + 7cosx - 1 = 0

6cos²x - 7cosx + 1 = 0

cosx = t

6t² - 7t + 1 = 0

√D = 5

t₁ = (7 - 5) / 12 = 1/6

t₂ = (7 + 5) / 12 = 1

cosx = 1/6

cosx = 1

x = arccos (1/6) + 2πn, где n ∈ Z

x = - arccos (1/6) + 2πn, где n ∈ Z

x = 2πn, где n ∈ Z

На отрезке [-3π; π] x равен: 0; - 2π; - arccos (1/6) - 2π; arccos (1/6) - 2π; - arccos (1/6) ; arccos (1/6)