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beatrisp
@beatrisp
August 2022
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(cosx-1/2) / (sinx-корень из 3/2) = 0
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hovsep50
(cosx-1/2) / (sinx-√3/2) = 0 ⇒
⇒ 1) cosx - 1/2 = 0 & 2) sinx - √3/2 ≠ 0
1) cosx = 1/2 ⇒ x = +/- π/3 +2πk ; k ∈Z
2) sinx ≠ √3/2 ⇒ x ≠ (-1)^n · π/3 + πn ; n ∈Z
Из 1) и 2) ⇒ x = - π/3 + 2πk ; k ∈Z
Ответ: x = 2πk - π/3 ; k∈Z
Более наглядно:
1) x = - π/3 + 2πk и x = π/3 + 2πk или x = {-π/3+2πk; π/3+2πk}
2) x ≠ - π/3 +(2k+1)·π и x ≠ π/3 + 2πk
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Answers & Comments
⇒ 1) cosx - 1/2 = 0 & 2) sinx - √3/2 ≠ 0
1) cosx = 1/2 ⇒ x = +/- π/3 +2πk ; k ∈Z
2) sinx ≠ √3/2 ⇒ x ≠ (-1)^n · π/3 + πn ; n ∈Z
Из 1) и 2) ⇒ x = - π/3 + 2πk ; k ∈Z
Ответ: x = 2πk - π/3 ; k∈Z
Более наглядно:
1) x = - π/3 + 2πk и x = π/3 + 2πk или x = {-π/3+2πk; π/3+2πk}
2) x ≠ - π/3 +(2k+1)·π и x ≠ π/3 + 2πk