Прежде чем приступить к решению примеров и задач, обязательно ознакомьтесь с теоретической частью урока
СИСТЕМЫ ТРИГОНОМЕТРИЧЕСКИХ УРАВНЕНИЙ
или посмотрите
ВИДЕО УРОК
1. Решить систему уравнений:
y =
±π/3 + 2πk, n, k ∈ Z;
б) x =
(–1)n ∙ π/6 + 2πn,
y =
±π/3 + πk, n, k ∈ Z;
в) x =
(–1)n ∙ π/6 + πn,
y =
±π/3 + 2πk, n, k ∈ Z;
г) x =
(–1)n ∙ π/6 + πn,
y =
±π/3 + πk, n, k ∈ Z.
y =
π/4 +
π(2k +
1),
m, k ∈ Z;
б) x =
–π/2 + πm,
y =
π/4 +
π(2k +
1),
m, k ∈ Z;
в) x = –π/2 + 2πm,
y =
π/4 +
π(2k +
1),
m, k ∈ Z;
г) x =
–π/2 + πm,
y =
π/4 +
π(k +
1),
m, k ∈ Z.
y1 =
π/6 –
2πn.
x2 =
–π/3 +
πn,
y2 =
5π/6 – 2πn;
б) x1 = π/3 + 2πn,
y1 =
π/6 –
πn.
x2 =
–π/3 +
2πn,
y2 =
5π/6 – πn;
в) x1 = π/3 + πn,
y1 =
π/6 –
πn.
x2 =
–π/3 +
πn,
y2 =
5π/6 – πn;
г) x1 = π/3 + 2πn,
y1 =
π/6 –
2πn.
x2 =
–π/3 +
2πn,
y2 = 5π/6 – 2πn.
y1 =
π/3 +
πn –
πk.
x2 =
–π/6 +
πn +
πk,
y2 =
–π/3 +
πn – πk;
б) x1 = π/3 + πn + πk,
y1 =
π/3 +
πn –
πk.
x2 =
– π/3 +
πn +
πk,
y2 =
–π/3 +
πn – πk;
в) x1 = π/6 + πn + πk,
y1 =
π/6 +
πn –
πk.
x2 =
–π/6 +
πn +
πk,
y2 =
–π/6 +
πn – πk;
г) x1 = π/3 + πn + πk,
y1 =
π/3 +
πn –
πk.
x2 =
–π/6 +
πn +
πk,
y2 =
–π/6 +
πn – πk.
y1 =
5π/6 – 2πn.
x2 =
5π/6 + πn,
y2 =
π/6 – 2πn;
б) x1 = π/6 + πn,
y1 =
5π/6 – πn.
x2 =
5π/6 + πn,
y2 = π/6 – πn;
в) x1 = π/6 + 2πn,
y1 =
5π/6 – πn.
x2 =
5π/6 + 2πn,
y2 =
π/6 – πn;
г) x1 = π/6 + 2πn,
y1 = 5π/6 – 2πn.
x2 =
5π/6 + 2πn,
y2 =
π/6 – 2πn.
y =
–π/3 +
2πn, n ∈ Z;
б) x =
π/3 + 2πn,
y =
–π/3 +
πn, n ∈ Z;
в) x =
π/3 + πn,
y =
–π/3 +
πn, n ∈ Z;
г) x =
π/3 + πn,
y =
–π/3 +
2πn, n ∈ Z.
y =
π/2 +
πn, k, n ∈ Z;
б) x =
π/2 + 2πk,
y =
π/2 +
2πn, k, n ∈ Z;
в) x =
π/2 + πk,
y =
π/2 +
πn, k, n ∈ Z;
г) x =
π/2 + πk,
y =
π/2 +
2πn, k, n ∈ Z.
y1 = (–1)n ∙ π/6 + πn.
x2 = (–1)k arcsin 2/3 + 2πk,
y2
= (–1)n arcsin 1/6 + πn;
б) x1
= πk,
y1
= (–1)n ∙ π/6 + 2πn.
x2
= (–1)k arcsin 2/3 + πk,
y2
= (–1)n arcsin 1/6 + 2πn;
в) x1
= πk,
y1
= (–1)n ∙ π/6 + πn.
x2
= (–1)k arcsin 2/3 + πk,
y2
= (–1)n arcsin 1/6 + πn;
г) x1
= 2πk,
y1
= (–1)n ∙ π/6 + 2πn.
x2
= (–1)k arcsin 2/3 + 2πk,
y2 = (–1)n arcsin 1/6 + 2πn.
y1 =
π/6 +
π(k – n).
x2 =
2π/3 + π(k +
n),
y2 =
–π/6 +
π(k – n);
б) x1 = π/3 + 2π(k +
n),
y1 =
π/6 +
2π(k –
n).
x2 =
2π/3 + π(k +
n),
y2 =
–π/6 +
π(k – n);
в) x1 = π/3 + 2π(k +
n),
y1 =
π/6 +
2π(k –
n).
x2 =
2π/3 + 2π(k +
n),
y2 =
–π/6 +
2π(k –
n);
г) x1 = π/3 + π(k +
n),
y1 =
π/6 +
π(k – n).
x2 =
2π/3 + 2π(k +
n),
y2 =
–π/6 +
2π(k –
n).
y1 =
π/4 +
2πn.
x2 =
–π/4 + 2πk,
y2 =
–3π/4 + 2πn;
б) x1 = π/4 + πk,
y1 =
π/4 +
πn.
x2 =
–π/4 + πk,
y2 =
–3π/4 + πn;
в) x1 = π/4 + 2πk,
y1 =
π/4 +
πn.
x2 =
–π/4 + 2πk,
y2 =
–3π/4 + πn;
г) x1 = π/4 + πk,
y1 =
π/4 +
2πn.
x2 =
–π/4 + πk,
y2 =
–3π/4 + 2πn.
y =
±2π/3 + 4πk +
2πn;
б) x = ±π/3 + 2πk,
y =
±2π/3 + 4πk +
2πn;
в) x =
±π/3 + πk,
y =
±2π/3 + 4πk +
πn;
г) x =
±π/3 + 2πk,
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