Решить уравнение 1+2sin2x+2cos ^2x=0

1 + 2sin2x + 2cos²x = 0

sin²x + cos²x + 4sinxcosx + 2cos²x = 0

sin²x + 4sinxcosx + 3cos²x = 0 |:cos²x

tg²x + 4tx + 3 = 0

tg²x + 4tgx + 4 - 1 = 0

(tgx + 2) ² - 1² = 0

(tgx + 2 - 1) (tgx + 2 + 1) = 0

(tgx - 1) (tgx + 3) = 0

1) tgx - 1 = 0

tgx = 1

x = π/4 + πn, n ∈ Z

2) tgx + 3 = 0

tgx = - 3

x = arctg (-3) + πk, k ∈ Z

Ответ: x = π/4 + πn, n ∈ Z; arctg (-3) + πk, k ∈ Z.