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СИСТЕМЫ ТРИГОНОМЕТРИЧЕСКИХ УРАВНЕНИЙ
или посмотрите
ВИДЕО УРОК
1. Решить систему уравнений:
y = π/6 – 2πk;
б) x =
π/6 –
2πk,
y = π/6 – 2πk;
в) x =
π/6 +
2πk,
y = π/6 + 2πk;
г) x =
π/6 –
2πk,
y = π/6 + 2πk.
y1 =
π/6 +
π(k +
n).
x2 =
– π/6 + π(k – m),
y2 =
2π/3 + π(k + m);
б) x1 = π/6 + π(k + n),
y1 =
π/3 +
π(k +
n).
x2 =
– π/6 + π(k + m),
y2 =
2π/3 + π(k + m);
в) x1 = π/6 + π(k – n),
y1 =
π/3 +
π(k +
n).
x2 =
– π/6 + π(k – m),
y2 =
2π/3 + π(k + m);
г) x1 = π/6 + π(k – n),
y1 =
π/3 +
π(k – n).
x2 =
– π/6 + π(k – m),
y2 =
2π/3 + π(k – m).
y = π/4 + (–1)n+1
∙ π/12 + πn/2 – πk;
б) x =
–π/4 + (–1)n+1
∙ π/12 + πn/2 + πk,
y = π/4 + (–1)n+1
∙ π/12 + πn/2 – πk;
в) x =
π/4 + (–1)n+1
∙ π/12 + πn/2 + πk,
y = –π/4 + (–1)n+1
∙ π/12 + πn/2 – πk;
г) x =
–π/4 + (–1)n+1
∙ π/12 + πn/2 + πk,
y = –π/4 + (–1)n+1 ∙ π/12 + πn/2 – πk.
y1
= π/4 – πk.
x2
= 3π/4 + πk,
y2 = – π/4 – πk, k ∈
Z;
б) x1
= π/4 + πk,
y1
= π/4 – πk.
x2
= 3π/4 + 2πk,
y2 = – π/4 – 2πk, k ∈
Z;
в) x1
= π/4 + 2πk,
y1
= π/4 – 2πk.
x2
= 3π/4 + πk,
y2 = – π/4 – πk, k ∈
Z;
г) x1 = π/4 + 2πk,
y1 =
π/4 –
2πk.
x2 =
3π/4 + 2πk,
y2 = – π/4 – 2πk, k ∈ Z.
y = πn, n, k ∈
Z;
б) x =
±π/3 + πk,
y = 2πn, n, k ∈
Z;
в) x =
±π/3 + 2πk,
y = πn, n, k ∈
Z;
г) x =
±π/3 + 2πk,
y = 2πn, n, k ∈ Z.
y = π/2 + 2πn, n, k ∈
Z;
б) x =
π/2 + 2πk,
y = π/2 + πn, n, k ∈
Z;
в) x =
π/2 + πk,
y = π/2 + πn, n, k ∈
Z;
г) x =
π/2 + πk,
y = π/2 + 2πn, n, k ∈ Z.
y
= π/6 + π(k – n), n, k ∈
Z;
б) x =
– π/6 + π(k + n),
y
= π/6 + π(k – n), n, k ∈
Z;
в) x =
– π/6 + π(k + n),
y
= –π/6 + π(k – n), n, k ∈
Z;
г) x =
π/6 + π(k + n),
y =
–π/6 +
π(k – n), n, k ∈ Z.
y
= π/4 – 2πn, n ∈
Z;
б) x =
π/4 + πn,
y
= π/4 – 2πn, n ∈
Z;
в) x =
π/4 + πn,
y
= π/4 – πn, n ∈
Z;
г) x =
π/4 + 2πn,
y =
π/4 –
πn, n ∈ Z.
y
= π/2 + π(n – k), n, k ∈
Z;
б) x =
π/2 + π(k + n),
y
= π/2 + π(n – k), n, k ∈
Z;
в) x =
– π/2 + π(k + n),
y
= –π/2 + π(n – k), n, k ∈
Z;
г) x =
π/2 + π(k + n),
y =
–π/2 +
π(n – k), n, k ∈ Z.
y1 =
π/4 –
πn.
x2 =
π/4 +
2πk,
y2 = –πk, k, n ∈ Z;
б) x1 =
πn,
y1 =
π/4 –
πn.
x2 =
π/4 +
πk,
y2 = –πk, k, n ∈ Z;
в) x1 = πn,
y1 =
π/4 –
2πn.
x2 =
π/4 +
πk,
y2 = –2πk, k, n ∈ Z;
г) x1 = 2πn,
y1 =
π/4 –
2πn.
x2 =
π/4 +
2πk,
y2 = –2πk, k, n ∈ Z.
y
= π/4 + πn/2 – πk, n, k ∈
Z;
б) x =
π/4 + πn/2.
y
= π/4 + πn/2 + 2πk, n,
k ∈ Z;
в) x =
π/4 +
πn/2.
y =
π/4 +
πn/2 –
2πk, n, k ∈ Z;
г) x =
π/4 +
πn/2.
y =
π/4 +
πn/2 +
πk, n, k ∈ Z.
y =
π/6 +
πn/2 +
πk, n, k ∈ Z;
б) x =
π/2 +
πn/2.
y =
π/6 +
πn/2 +
πk, n, k ∈ Z;
в) x =
π/4 +
πn/2.
y =
π/3 +
πn/2 +
πk, n, k ∈ Z;
г) x =
π/2 +
πn/2.
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