§ 32: 32.7 (1)

32.7 Розв’яжіть рівняння:

1) sin²x2 + sin²3x2 = 1;

sin²3x2 + sin²x2 – 1 = 0; sin²3x2 – cos²x2 = 0;

(sin3x2 – cosx2)(sin3x2 + cosx2) = 0;

(sin3x2 – sin(π2 – x2)) • (sin3x2 + sin(π2 – x2)) = 0;

2sin(–π4 + x) • cos(π4 + x2) • 2sin(π4 + x2) • cos(–π4 + x) = 0;

2sin(–π2 + 2x) • sin(π2 + x) = 0;

2cos 2x • cos x = 0; cosx = 0; x = π2 + πn, n ∈ Z;

cos2x = 0; 2x = π2 + πk, k ∈ Z;

x = π2 + πn, n ∈ Z;

x = π4 + πk2, k ∈ Z.

Відповідь: x = π2 + πn, n ∈ Z; x = π4 + πk2, k ∈ Z.

2) cos2x – cos8x + cos6x = 1; cos2x + cos6x – 1 – cos8x = 0;

2cos4xcos(–2x) – 1 – 2cos²4x + 1 = 0;

2cos4x • cos2x – 2cos²4x = 0; 2cos4x(cos2x – 1) = 0;

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

cos2x = 1; 2x = 2πk, k ∈ Z; x = πk, k ∈ Z.

Відповідь: x = π8 + πn4, n ∈ Z; x = πk, k ∈ Z.

3) 1 – cosx = tgx – sinx;

1 – cosx = sinxcosx – sinx; cosx – cos²x = sinx – sinx • cosx;

cosx – sinx – cos²x + sinx • cosx = 0;

cosx – sinx – cosx(cosx – sinx) = 0;

(cosx – sinx)(1 – cosx) = 0;

cosx = 1; cosx = 1; cosx = 1; x = 2πn, n ∈ Z;

cosx – sinx = 0; 1 – tgx = 0; tgx = 1; x = arctg1 + πk, k ∈ Z;

x = 2πn, n ∈ Z;

x = π4 + πk, k ∈ Z.

Відповідь: x = 2πn, n ∈ Z; x = π4 + πk, k ∈ Z.

4) sinx + sin3x = 4cos²x; 2sin2x • cos(–x) – 4 cos²x = 0;

2sin2x • cosx – 4cos²x = 0; 4sinx • cos²x – 4 cos²x = 0;

4cos²x(sinx – 1) = 0;

cosx = 0; x = π2 + πn, n ∈ Z;

sinx = 0; x = π2 + πk, k ∈ Z; ⇒ x = π2 + πn, n ∈ Z.

Відповідь: x = π2 + πn, n ∈ Z.

5) сos2x = 2(cosx – sinx); cos2x – 2(cosx – sinx) = 0;

cos²x – sin²x – 2(cosx – sinx) = 0;

cos2x – sin2x – 2(cosx – sinx) = 0;

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

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

cosx – sinx = 0; 22 cos x – 22 sin x = 0; sin(π4 – x) = 0;

cosx + sinx = 2; 22 cos x + 22 sin x = 1; sin(π4 + x) = 1;

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

π4 + x = π2 + 2πk, k ∈ Z; x = π4 + 2πk, k ∈ Z; ⇒

⇒ x = π4 + πn, n ∈ Z.

Відповідь: x = π4 + πn, n ∈ Z.

6) sin3x + 3cos3x = 2cos5x|: 2 ⇒ 12sin3x + 32cos3x = cos5x;

cosπ6cos3x + sinπ6 • sin3x = cos5x; cos(π6 – 3x) – cos5x = 0;

–2sin(π12 + x) • sin(π12 – 4x) = 0; 2sin(x + π12) • sin(4x – π12) = 0;

sin(x + π12) = 0; x + π12 = πn, n ∈ Z; x = –π12 + πn, n ∈ Z;

sin(4x – π12) = 0; 4x – π12 = πk, k ∈ Z; 4x = π12 + πk, k ∈ Z;

x = –π12 + πn, n ∈ Z;

x = π48 + πk4, k ∈ Z.

Відповідь: x = –π12 + πn, n ∈ Z; x = π48 + πk4, k ∈ Z.