task/29729177 Решить уравнение ctg(2x) - ctg(x) = 2ctg(4x) ----------
ОДЗ : { sin2x ≠ 0 ; sinx ≠ 0 ; sin4x ≠0 . x ≠ πk/4 , k ∈ ℤ .
ctg(2x) - ctg(x) = 2ctg(4x) ⇔ ctg(2x) - 2ctg(4x) = ctg(x) ⇔
ctg(2x) -(ctg²(2x)-1) /ctg2x =ctg(x) ⇔1/ctg(2x)=ctg(x)⇔2ctgx / (ctg²x -1) =ctgx⇔
|| ctgx ≠ 0 || 2 / (ctg²x -1) = 1 ⇔ 2 = ctg²x - 1 ⇔ ctg²x = 3 ⇔ || ctgx = ±√3 ||
(1+cos2x) / (1-cos2x) = 3 ⇔ 1+cos2x =3 - 3cos2x ⇔ cos2x = 1/2 ⇔
2x = ± π/3 + 2πk , k ∈ ℤ .
ответ: x =± π/6 + πk , k ∈ ℤ
Copyright © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
Verified answer
Verified answer
task/29729177 Решить уравнение ctg(2x) - ctg(x) = 2ctg(4x) ----------
ОДЗ : { sin2x ≠ 0 ; sinx ≠ 0 ; sin4x ≠0 . x ≠ πk/4 , k ∈ ℤ .
ctg(2x) - ctg(x) = 2ctg(4x) ⇔ ctg(2x) - 2ctg(4x) = ctg(x) ⇔
ctg(2x) -(ctg²(2x)-1) /ctg2x =ctg(x) ⇔1/ctg(2x)=ctg(x)⇔2ctgx / (ctg²x -1) =ctgx⇔
|| ctgx ≠ 0 || 2 / (ctg²x -1) = 1 ⇔ 2 = ctg²x - 1 ⇔ ctg²x = 3 ⇔ || ctgx = ±√3 ||
(1+cos2x) / (1-cos2x) = 3 ⇔ 1+cos2x =3 - 3cos2x ⇔ cos2x = 1/2 ⇔
2x = ± π/3 + 2πk , k ∈ ℤ .
ответ: x =± π/6 + πk , k ∈ ℤ