ПОМОГИТЕ!!!!!!!!!!! cos2x+2=0sin4x=02sin x/2+1=02cos2x-1=02tg^2x-tgx=0

cos2x + 2 = 0 cos2x = -2 Нет корней, т.к. косинус аргумента принадлежит отрезку [-1; 1]. sin4x = 0 4x = πn, n ∈ Z x = πn/4, n ∈ Z. 2sin(x/2) + 1 = 0 sin(x/2) = -1/2 x/2 = (-1)ⁿ+¹π/6 + πn, n ∈ Z x = (-1)ⁿ+¹π/3 + πn, n ∈ Z. 2cos2x - 1 = 0 cos2x = 1/2 2x = ±π/3 + 2πn, n ∈ Z. x = ±π/6 + πn, n ∈ Z 2tg²x - tgx = 0 tgx(2tgx - 1) = 0 tgx = 0 x = πn, n ∈ Z. 2tgx - 1 = 0 tgx = 1/2 x = arctg(1/2) + πn, n ∈ Z.