Question

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

Answer 1

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.