§ 32: 32.5 (1-2)

32.5 Розв’яжіть рівняння:

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

sinx = 12; x=(–1)k • arcsin12 + πk, k ∈ Z; x = (–1)^k • π6 + πk, k ∈ Z;

cosx = –1; x = – π + 2πn, n ∈ Z; x = –π + 2πn, n ∈ Z.

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

2) 1 + cos3x = cos4x; 1 + 2cos(2•4x) – 2 – cos4x = 0; cos4x(2cos

4x – 1) = 0;

cos4x = 0; 4x = π2 + πn, n ∈ Z; x = π8 + πn4, n ∈ Z;

cos 4x = 12; 4x = ±arccos 12 + 2πk, k ∈ Z; x = ± π12 + πk2, k ∈ Z.

Відповідь: x = π8 + πn4, n ∈ Z; x = ± π12 + πk2, k ∈ Z.

3) 2sin2x + cos3x – cosx = 0; 2sin2x – 2 • sin2x • sinx = 0;

2sin2x(1 – sinx) = 0;

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

sinx = 1; x = π2 + 2πk, k ∈ Z; x = π2 + 2πk, k ∈ Z. ⇒

⇒ x = πn2, n ∈ Z.

Відповідь: x = πn2, n ∈ Z.

4) sin4x + 2cos²x = 1; 2sin2x • cos2x + 2cos²x – 1 = 0; 2sin2x • cos2x + cos2x = 0; cos2x • (2sin2x + 1) = 0;

cos 2x = 0; 2x = π2 + 2πn, n ∈ Z.

sin 2x = –12; 2x = (–1)^k • arccos(–12) + πk, k ∈ Z.

x = π4 + πn, n ∈ Z; x = π4 + πn, n ∈ Z;

2x = (–1)^k+1 • π2 + 2πk, k ∈ Z; x = (–1)^k+1 • π12 + πk2, k ∈ Z;

Відповідь: x = π4 + πn, n ∈ Z; x = (–1)^k+1 • π12 + πk2, k ∈ Z.

5) cosx – cos3x = 3sin²x; –2sin2x • sin(–x) = 3sin²x;

2sin2x • sinx – 3sin²x = 0; sinx(2sin2x – 3sinx) = 0;

sinx • (4sinx • cosx – 3sinx) = 0; sin²x(4cosx – 3) = 0;

sinx = 0; x = πn, n ∈ Z;

cosx = 34; x = ± arccos34 + 2πk, k ∈ Z.

Відповідь: x = πn, n ∈ Z; x = ± arccos34 + 2πk, k ∈ Z.

6) sinx + sin2x + sin3x + sin4x = 0; sinx + sin4x + sin2x + sin3x = 0; 2sin 5x2 • cos3x2 + 2sin5x2 • cosx2 = 0; 2sin5x2 • (cos3x2 + cosx2) = 0; 2sin5x2 • 2 cos x • cosx2 = 0;

sin5x2 = 0; 5x2 = πn, n ∈ Z; x = 2πn5, n ∈ Z;

cosx = 0; x = π2 + 2πk, k ∈ Z; x = π2 + 2πk, k ∈ Z;

cosx2 = 0; x2 = π2 + 2πl, l ∈ Z; x = π + 4πl, l ∈ Z.

Відповідь: x = 2πn5, n ∈ Z; x = π2 + 2πk, k ∈ Z; x = π + 4πl, l ∈ Z.