sin4x + sin2x = 0
sin2x = 2•sinx•cosx - синус двойного аргумента
2•sin2x•cos2x + sin2x = 0
sin2x•(2cos2x + 1) = 0
1) sin2x = 0 ⇔ 2x = πn ⇔ x = πn/2, n ∈ Z
2) 2cos2x + 1 = 0 ⇔ cos2x = - 1/2 ⇔ 2x = (± 2π/3) + 2πk ⇔ x = (± π/3) + πk, k ∈ Z
ОТВЕТ: πn/2, n ∈ Z ; (± π/3) + πk, k ∈ Z