X₁ = arctg( - 2√2) + πk, k ∈ ZРешение
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