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

cos2x + 2 = 0cos2x = -2Нет корней, т.к. косинус аргумента принадлежит отрезку [-1; 1].sin4x = 04x = πn, n ∈ Zx = πn/4, n ∈ Z.2sin(x/2) + 1 = 0sin(x/2) = -1/2x/2 = (-1)ⁿ+¹π/6 + πn, n ∈ Zx = (-1)ⁿ+¹π/3 + πn, n ∈ Z.2cos2x - 1 = 0cos2x = 1/22x = ±π/3 + 2πn, n ∈ Z.x = ±π/6 + πn, n ∈ Z 2tg²x - tgx = 0tgx(2tgx - 1) = 0tgx = 0x = πn, n ∈ Z.2tgx - 1 = 0tgx = 1/2x = arctg(1/2) + πn, n ∈ Z.