1/tg²x + 9/tgx + 8 = 0
ctg²x + 9ctgx + 8 = 0
Пусть t = ctgx.
t² + 9t + 8 = 0
t1 + t2 = 9
t1•t2 = 8
t1 = 8
t2 = 1
Обратная замена:
ctgx = 8
x = arcctg8 + πn, n ∈ Z
ctgx = 1
x = π/4 + πn, n ∈ Z
3π/2 < π/4 + πn < 5π/2
6π < π + 4πn < 10π
6 < 1 + 4n < 10
5 < 4n < 9
При n ∈ Z n = 2.
π/4 + 2π = 9π/4.
Ответ: 9π/4; arcctg8 + 2π.