Решите уравнение cos (2 П-2x) + 3sin (П-x) = 2

Решение

cos (2 П-2x) + 3sin (П-x) = 2

cos2x + 3sinx = 2

1 - 2sin²x + 3sinx - 2 = 0

2sin²x - 3sinx + 1 = 0

sinx = t

2t² - 3t + 1 = 0

D = 9 - 4*2*1 = 1

t₁ = (3 - 1) / 4

t₁ = 1/2

t₂ = (3 + 1) / 4

t₂ = 1

1) sinx = 1/2

x = (-1) ^n * arcsin (1/2) + πk, k∈Z

x₁ = (-1) ^n * (π/6) + πk, k∈Z

2) sinx = 1

x₂ = π/2 + πn, n∈Z