Решение
4tg5x = √3(2 - 2tg²5x)
2√3 - 2√3tg²5x - 4tg5x = 0
2√3tg²(5x) + 4tg(5x) - 2√3 = 0
tg(5x) = t
2√3t² + 4t - 2√3 = 0
D = 16 + 4*(2√3)*(2√3) = 64
t₁ = (-4 - 8)/4√3
t₁ = - 3/√3 = - √3
t₂ = (-4 + 8)/4√3
t₂ = 1/√3
1) tg(5x) = - √3
5x = arctg( - √3) + πk, k∈Z
5x = - π/3 + πk, k∈Z
x₁ = - π/15+ πk/5, k∈Z
2) tg(5x) = 1/√3
5x = arctg(1/√3) + πn, n∈Z
5x = π/6+ πn, n∈Z
x₂ = π/30 + πn/5, n∈Z
Ответ: Е)
x₁ = - π/15+ πk/5, k∈Z
x₂ = π/30 + πn/5, n∈Z