1/(cos^4)x-(tg^4)x=17

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