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VikaMakarskaya
@VikaMakarskaya
July 2022
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1/cos^2x=3+tgx
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ОльгаРоманова99
1/соs²x = 3 + tg x
tg²x + 1 = 3 + tg x
tg²x - tg x + 1 - 3 = 0
tg²x - tg x - 2 = 0
★ Пусть tg x = a, тогда:
а² - а - 2 = 0
По теореме обратной теореме Виета:
а1 × а2 = -2 ; а1 + а2 = 1 => а1 = 2 ; а2 = -1
★ tg x = 2 или tg x = -1
x1 = arctg 2 + πk, k [принадлежит] Z ;
x2 = -π/4 + πn, n [принадлежит] Z
Ответ: arctg 2 + πk, k [принадлежит] Z ; -π/4 + πn, n [принадлежит] Z
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Answers & Comments
tg²x + 1 = 3 + tg x
tg²x - tg x + 1 - 3 = 0
tg²x - tg x - 2 = 0
★ Пусть tg x = a, тогда:
а² - а - 2 = 0
По теореме обратной теореме Виета:
а1 × а2 = -2 ; а1 + а2 = 1 => а1 = 2 ; а2 = -1
★ tg x = 2 или tg x = -1
x1 = arctg 2 + πk, k [принадлежит] Z ;
x2 = -π/4 + πn, n [принадлежит] Z
Ответ: arctg 2 + πk, k [принадлежит] Z ; -π/4 + πn, n [принадлежит] Z