Тиньков
5 год назад
Tg^2x+4tgx-5 = 0, -pi/2<x<0; ctg2x = ?
ОТВЕТЫ
Харламович
Jun 30, 2019
Tgx = a
a² + 4a - 5 = 0
D = 16 + 20 = 36
a = -5 или a = 1
tgx = -5 tgx = 1
x = -arctg 5 + πn x = π/4 + πk
x = - arctg 5 ∈ ( - π/2; 0) x∉ ( - π/2 ; 0)
x = - arctg 5 , т.е. tg x = - 5 ⇒ ctg x = - 1/5
ctg2x = (ctg²x - 1)/ (2 ctg x) = (1/25 - 1) / (-2/5) = 24/25 · 5/2 = 12/5
a² + 4a - 5 = 0
D = 16 + 20 = 36
a = -5 или a = 1
tgx = -5 tgx = 1
x = -arctg 5 + πn x = π/4 + πk
x = - arctg 5 ∈ ( - π/2; 0) x∉ ( - π/2 ; 0)
x = - arctg 5 , т.е. tg x = - 5 ⇒ ctg x = - 1/5
ctg2x = (ctg²x - 1)/ (2 ctg x) = (1/25 - 1) / (-2/5) = 24/25 · 5/2 = 12/5
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