1) 23cosx+19sin^2x+19cos^2x=18 2) cos^2 * 2x/23=3/4 3) 17sin^2x+289sinx=0

1)

23cosx + 19sin²x + 19cos²x = 18; 23cosx + 19 (sin²x + cos²x) = 18; 23cosx + 19 = 18; 23cosx = 18 - 19; 23cosx = - 1; cosx = - 1/23; x = π ± arccos (1/23) + 2πk, k ∈ Z.

2)

cos² (2x/23) = 3/4; cos (2x/23) = ±√3/2; 2x/23 = ±π/6 + πk, k ∈ Z; 2x = ±23π/6 + 23πk, k ∈ Z; x = ±23π/12 + 23πk/2, k ∈ Z.

3)

17sin²x + 289sinx = 0; 17sinx (sinx + 17) = 0; [17sinx = 0;

[sinx + 17 = 0; [sinx = 0;

[sinx = - 17 < - 1 - не имеет решения. x = πk, k ∈ Z.

Ответ:

1) π ± arccos (1/23) + 2πk; 2) ±23π/12 + 23πk/2; 3) πk, k ∈ Z.