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UAGamesTactics
@UAGamesTactics
July 2022
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Помогите решить тригонометрию
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Dимасuk
Verified answer
1) 6sin²x - cosx + 6 = 0
6 - 6cos²x - cosx + 6 = 0
-6cos²x - cosx + 12 = 0
6cos²x + cosx - 12 = 0
Пусть t = cosx, t ∈ [-1; 1]
D = 1 + 12·4·6 = 289 = 17²
t₁ = (-1 + 17)/12 = 16/12 - посторонний корень
t₂ = (-1 - 17)/12 = -18/12 - посторонний корень
Ответ: нет корней.
2) tg²x + 3tgx = 0
tgx(tgx + 3) = 0
tgx = 0 или tgx = -3
x = πn, n ∈ Z или x = arctg(-3) + πk, k ∈ Z
Ответ: x = πn, n ∈ Z; arctg(-3) + πk, k ∈ Z .
3) 2sin2x = 3cos2x |:cos2x
2tg2x = 3
tg2x = 3/2
2x = arctg(3/2) + πn, n ∈ Z
x = 1/2·arctg(3/2) + πn/2, n ∈ Z
Ответ: x = 1/2·arctg(3/2) + πn/2, n ∈ Z.
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Answers & Comments
Verified answer
1) 6sin²x - cosx + 6 = 06 - 6cos²x - cosx + 6 = 0
-6cos²x - cosx + 12 = 0
6cos²x + cosx - 12 = 0
Пусть t = cosx, t ∈ [-1; 1]
D = 1 + 12·4·6 = 289 = 17²
t₁ = (-1 + 17)/12 = 16/12 - посторонний корень
t₂ = (-1 - 17)/12 = -18/12 - посторонний корень
Ответ: нет корней.
2) tg²x + 3tgx = 0
tgx(tgx + 3) = 0
tgx = 0 или tgx = -3
x = πn, n ∈ Z или x = arctg(-3) + πk, k ∈ Z
Ответ: x = πn, n ∈ Z; arctg(-3) + πk, k ∈ Z .
3) 2sin2x = 3cos2x |:cos2x
2tg2x = 3
tg2x = 3/2
2x = arctg(3/2) + πn, n ∈ Z
x = 1/2·arctg(3/2) + πn/2, n ∈ Z
Ответ: x = 1/2·arctg(3/2) + πn/2, n ∈ Z.