§ 32: 32.6 (1-2)

32.6 Розв’яжіть рівняння:

1) sin2x + 2sinx = 0; 2sinx • cosx + sinx = 0;

sinx(2cosx + 1) = 0;

sinx = 0; x = πn, n ∈ Z; x = πn, n ∈ Z;

cosx = –12; x = ±arcsin(–12) + πk, k ∈ Z; x = ± 2π3 + πk, k ∈ Z.

Відповідь: x = πn, n ∈ Z; x = ± 2π3 + πk, k ∈ Z.

2) sin2x – cosx = 2sinx – 1; 2sinx • cosx – cosx = 2sinx – 1;

cosx(2sinx – 1) = 0; (2sinx – 1)(cosx – 1) = 0;

cosx = 0; x = 2πn, n ∈ Z;

sinx = 12; x = (–1)^k • arcsin 12 + πk, k ∈ Z.

x = 2πn, n ∈ Z;

x = (–1)^k • π6 + πk, k ∈ Z.

Відповідь: x = 2πn, n ∈ Z; x = (–1)^k • π6 + πk, k ∈ Z.

3) 1 – cos8x = sin4x; 2sin²4x = sin4x; sin4x(2sin4x – 1) = 0;

sin4x = 0; 4x = πn, n ∈ Z;

sin4x = 12; 4x = (–1)^k • arcsin12 + πk, k ∈ Z.

x = πn4, n ∈ Z; x = πn4, n ∈ Z;

4x = (–1)^k • π6 + πk, k ∈ Z. x = (–1)^k • π24 + πk4, k ∈ Z.

Відповідь: x = πn4, n ∈ Z; x = (–1)^k • π24 + πk4, k ∈ Z.

4) sinx + sin2x + sin3x = 0; sinx + sin3x + sin2x = 0;

2sin2x • cos(–x) + sin2x = 0; sin2x(2cosx + 1) = 0;

sin2x = 0; 2x = πn, n ∈ Z; x = πn2, n ∈ Z;

cosx = –12; x = ± arccos(–12) + 2πk, k ∈ Z; x = ± 2π3 + 2πk, k ∈ Z.

Відповідь: x = πn2, n ∈ Z; x = ± 2π3 + 2πk, k ∈ Z.

5) cos9x – cos7x + cos3x – cosx = 0; –2sin8x • sinx – 2sin2x • sinx = 0; –2sinx(sin8x + sin2x) = 0; –2sin x • 2sin5x • cos3x = 0;

sinx = 0; x = πn, n ∈ Z; x = πn, n ∈ Z;

sin5x = 0; 5x = πk, k ∈ Z; x = πk5, k ∈ Z;

cos3x = 0; 3x = π2 + πl, l ∈ Z; x = π6 + πl3, l ∈ Z.

Відповідь: x = πn, x = πk5; x = π6 + πl3, n, k, l ∈ Z.

6) 2cos5x + sin5x + sin3x – sin7x = 0; 2cos5x + sin(–2x) • cos5x = 0; cosx • (2 – 2sin2x) = 0;

cos5x = 0; 5x = π2 + πn, n ∈ Z;

sin2x = 22; 2x = (–1)^k • arcsin22 + πk, k ∈ Z;

x = π10 + πn5, n ∈ Z;

x = (–1)^k • π8 + πk2, k ∈ Z.

Відповідь: x = π10 + πn5, n ∈ Z; x = (–1)^k • π8 + πk2, k ∈ Z.