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)