Решение
tg((п/2)+x)-tg(2п-x)=(2√3)/3
- ctgx + tgx = (2√3)/3
1/tgx - tgx = (2√3)/3
3tg²x + 2√3tgx - 3 = 0
tgx = t
3t² + 2√3t - 3 = 0
D = 12 + 4*3*3 = 48
t₁ = (- 2√3 - 4√3)/6
t₁ = - 6√3 / 6
t₁ = - √3
t₂ = (- 2√3 +4√3)/6
t₂ = 2√3 / 6
t₂ = √3/3
1) tgx = - √3
x = arctg(- √3) + πk, k ∈ Z
x₁ = - π/3 + πk, k ∈ Z
2) tgx = √3/3
x = arctg(√3/3) + πn, n ∈ Z
x₂ = π/6 + πn, n ∈ Z
Ответ: x₁ = - π/3 + πk, k ∈ Z ; x₂ = π/6 + πn, n ∈ Z