§ 32: 32.17 (1)

32.17 Розв’яжіть рівняння:

1) sin7x – 2cos5x + 3cos7x – 2sin5x = 0| • 12 ⇒

12sin7x + 32cos7x – (22 • cos5x + 22 • sin5x) = 0;

(cosπ6 • cos7x + sinπ6 • sin7x) – (cosπ4 • cos5x + sinπ4 • sin5x) = 0;

cos(7x – π6) – cos(5x – π4) = 0; –2sin(6x – 5π24) • sin(x + π24) = 0;

sin(6x – 5π24) = 0; 6x – 5π24 = πn, n ∈ Z; 6x = 5π24 + πn, n ∈ Z;

sin(x + π24) = 0; x + π24 = πk, k ∈ Z; x = –π24 + πk, k ∈ Z;

x = 5π144 + πn, n ∈ Z;

x = –π24 + πk, k ∈ Z.

Відповідь: x = 5π144 + πn, n ∈ Z; x = –π24 + πk, k ∈ Z.

2) 2sin3x + sinx – cos2x = 3(sin2x – cosx);

2sin3x + sinx + 3cosx – (cos2x + 3sin2x) = 0| • 12 ⇒

sin3x + (12sinx + 32cosx) – (12cos3x + 32sin2x) = 0;

sin3x + (sinx • cosπ3 + cosx • sinπ3) – (cos2x • sinπ6 + sin2x • cos π6) = 0;

sin3x + sin(x + π3) – sin(2x + π6) = 0; 2sin(2x + π6) = 0;

sin3x – sin(2x + π6) = 0; sin(2x + π6) • (2cos(x – π6) – 1) = 0;

sin(2x + π6) = 0; 2x + π6 = πn, n ∈ Z;

cos(x – π6) = 12; x – π6 = ± π3 + 2πk, k ∈ Z;

2x = π6 = πn, n ∈ Z; x = π12 + πn2, n ∈ Z.

x = π6 ± π3 + 2πk, k ∈ Z; x = π6 ± π3 + 2πk, k ∈ Z.

Відповідь: x = π12 + πn2, n ∈ Z; x = π6 ± π3 + 2πk, k ∈ Z.

3) 3(2 – cosx) + 4sin2x = sinx; 23 + 4sin2x – (3cosx + sinx) = 0; 4(32 + sin2x) – 2(32cosx + 12sinx) = 0;

4(sinπ3 + sin2x) – 2(cos π6 • cos x + sinπ6 • sinx) = 0;

4 • 2sin(x + π6) • cos(π6 – x) – 2(cosx – π6) = 0|: 2 ⇒

4sin(x + π6) • cos(x – π6) – cos(x – π6) = 0;

cos(x – π6)(4sin(x + π6) – 1) = 0;

cos(x – π6) = 0; x – π6 = π2 + πn, n ∈ Z;

4sin(x + π6) – 1 = 0; sin(x + π6) = 14;

x = π6 + π2 + πn, n ∈ Z;

x + π6 = (–1)^k • arcsin 14 + πk, k ∈ Z;

x + 2π3 + πn, n ∈ Z;

x = –π6 + (–1)^k • arcsin 14+πk, k ∈ Z.

Відповідь: x + 2π3 + πn, n ∈ Z; x = –π6 + (–1)^k • arcsin 14 + πk, k ∈ Z.