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.