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СИСТЕМИ ТРИГОНОМЕТРИЧНИХ РІВНЯНЬ
або
ВИДЕО УРОК
1. Розв’яжіть систему рівнянь:
y =
±2π/3 + 2πk, n, k ∈ Z;
б) x = (–1)n ∙ π/6 + πn.
y =
±2π/3 + 2πk, n, k ∈ Z;
в) x = (–1)n ∙ π/6 + 2πn.
y =
±2π/3 + πk, n, k ∈ Z;
г) x = (–1)n ∙ π/6 + πn.
y =
±2π/3 + πk, n, k ∈ Z.
y =
±π/3 + (1 + 2n)π, n ∈ Z;
б) x =
±π/3 + πn.
y =
±π/3 + (1 + 2n)π, n ∈ Z;
в) x =
±π/3 + πn.
y =
±π/3 + (1 – 2n)π, n ∈ Z;
г) x =
±π/3 + 2πn.
y =
±π/3 + (1 – 2n)π, n ∈ Z.
y =
π/2(k – n), n, k ∈ Z;
б) x =
π/3(k +
n).
y =
π/2(k – n), n, k ∈ Z;
в) x =
π/2(k +
n).
y =
π/3(k – n), n, k ∈ Z;
г) x =
π/3(k +
n).
y =
π/3(k – n), n, k ∈ Z.
а) 1;
б) 0;
в) –1;
г) –0,5.
y
= π/2 + πk, n, k ∈
Z;
б) x =
(–1)n ∙ π/4 + 2πn.
y
= π/2 + πk, n, k ∈
Z;
в) x =
(–1)n ∙ π/4 + 2πn.
y
= π/2 + 2πk, n, k ∈
Z;
г) x =
(–1)n ∙ π/4 + πn.
y =
π/2 +
2πk, n, k ∈ Z.
y1
= 30° – 180°n.
x2
= 30° + 90°n,
y2 = 45° – 180°n, n ∈
Z;
б) x1
= 45° + 180°n,
y1
= 30° – 90°n.
x2
= 30° + 180°n,
y2 = 45° – 90°n, n ∈
Z;
в) x1
= 45° + 180°n,
y1
= 30° – 180°n.
x2
= 30° + 180°n,
y2 = 45° – 180°n, n ∈
Z;
г) x1
= 45° + 90°n,
y1 =
30° – 90°n.
x2 =
30° + 90°n,
y2 = 45° – 90°n, n ∈ Z.
y
= π/4 + πn/2 – πk, n, k ∈
Z;
б) x =
–π/4 + πn/2 + πk.
y
= π/4 + πn/2 – πk, n, k ∈
Z;
в) x =
–π/4 + πn/2 + πk.
y
= π/4 + πn/2 – 2πk, n, k ∈
Z;
г) x =
–π/4 + πn/2 + 2πk.
y =
π/4 +
πn/2 –
2πk, n, k ∈ Z.
y1
= π/2 ∓ π/3 + 2π(n – k).
x2
= –π/2 ± 2π/3 + π(n + k),
y2 = –π/2 ∓ 2π/3 + 2π(n – k), n, k ∈
Z;
б) x1
= π/2 ± π/3 + 2π(n + k),
y1
= π/2 ∓ π/3 + π(n – k).
x2
= –π/2 ± 2π/3 + 2π(n + k),
y2 = –π/2 ∓ 2π/3 + π(n – k), n, k ∈
Z;
в) x1
= π/2 ± π/3 + π(n + k),
y1
= π/2 ∓ π/3 + π(n – k).
x2
= –π/2 ± 2π/3 + π(n + k),
y2 = –π/2 ∓ 2π/3 + π(n – k), n, k ∈
Z;
г) x1 =
π/2 ± π/3 + 2π(n + k),
y1 =
π/2 ∓ π/3 + 2π(n – k).
x2 =
–π/2 ± 2π/3 + 2π(n + k),
y
= πn,
n, ∈ Z;
б) x =
π/2 – πn.
y
= πn,
n, ∈ Z;
в) x =
π/2 – πn.
y
= 2πn,
n, ∈ Z;
г) x =
π/2 – 2πn.
y =
2πn, n, ∈ Z.
y
= (–1)k+1 ∙ π/6 + π(1 – k),
k ∈ Z;
б) x =
(–1)k ∙ π/6 + πk.
y
= (–1)k+1 ∙ π/6 + 2π(1 – k),
k ∈ Z;
в) x =
(–1)k ∙ π/6 + 2πk.
y
= (–1)k+1 ∙ π/6 + 2π(1 – k),
k ∈ Z;
г) x =
(–1)k ∙ π/6 + πk.
y =
(–1)k+1
∙ π/6 + π(1
– k), k ∈ Z.
y1 =
–π/6 + π(n – k).
x2 =
–π/6 + π(n + k),
y2 = π/6 + π(n – k), n, k ∈ Z;
б) x1 = π/6 + 2π(n + k),
y1 =
–π/6 + π(n – k).
x2 =
–π/6 + 2π(n + k),
y2 = π/6 + π(n – k), n, k ∈ Z;
в) x1 = π/6 + 2π(n + k),
y1 =
–π/6 + 2π(n – k).
x2 =
–π/6 + 2π(n + k),
y2 = π/6 + 2π(n – k), n, k ∈ Z;
г) x1 = π/6 + π(n + k),
y1 =
–π/6 + 2π(n – k).
x2 =
–π/6 + π(n + k),
y2 = π/6 + 2π(n – k), n, k ∈ Z.
y =
π/3 –
2πn, n, ∈ Z;
б) x =
π/3 +
2πn.
y =
π/3 –
πn, n, ∈ Z;
в) x =
π/3 +
2πn.
y =
π/3 –
2πn, n, ∈ Z;
г) x =
π/3 +
πn.
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