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Dreymers
@Dreymers
July 2022
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ПОМОГИТЕ!!!!!!!!!!!
cos2x+2=0
sin4x=0
2sin x/2+1=0
2cos2x-1=0
2tg^2x-tgx=0
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Dимасuk
Verified answer
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.
2 votes
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Answers & Comments
Verified answer
Cos2x + 2 = 0cos2x = -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.