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makarevig
@makarevig
July 2022
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1/(cos^4)x-(tg^4)x=17
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nKrynka
X₁ = arctg( - 2√2) + πk, k ∈ ZРешение
1/(cos^4)x-(tg^4)x=17
(tg
²x + 1)² - tg⁴x - 17 = 0
tg⁴x + 2tg²x + 1
- tg⁴x - 17 = 0
2tg
²x = 16
tg²x = 8
1) tgx = 2√2
x₁ = arctg( 2√2) + πk, k ∈ Z
2) tgx = - 2√2
x₂ = arctg( - 2√2) +
πk, k ∈ Z
x₂ = - arctg( 2√2) +
πk, k ∈ Z
1 votes
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Answers & Comments
1/(cos^4)x-(tg^4)x=17
(tg²x + 1)² - tg⁴x - 17 = 0
tg⁴x + 2tg²x + 1 - tg⁴x - 17 = 0
2tg²x = 16
tg²x = 8
1) tgx = 2√2
x₁ = arctg( 2√2) + πk, k ∈ Z
2) tgx = - 2√2
x₂ = arctg( - 2√2) + πk, k ∈ Z
x₂ = - arctg( 2√2) + πk, k ∈ Z