1) вычислить Sin (p/48) * cos (p/48) * cos (p/24) * cos (p/12) 2) решите уравнение 26 sinx * cos x - cos4x+7=0 3) sin^2 (2x-p/6) = 3/4 4) sin^2 (x/2) = cos^2 (7x/2)

1)

A = sin (π/48) * cos (π/48) * cos (π/24) * cos (π/12); A = 1/2sin (π/24) * cos (π/24) * cos (π/12); A = 1/4sin (π/12) * cos (π/12); A = 1/8sin (π/6) = 1/8 * 1/2 = 1/16.

2)

26sinx * cosx - cos4x + 7 = 0; 13sin2x - (1 - 2sin^2 (2x)) + 7 = 0; 2sin^2 (2x) + 13sin2x + 6 = 0; D = 13^2 - 4 * 2 * 6 = 169 - 48 = 121; sin2x = (-13 ± √121) / 4 = (-13 ± 11) / 4; 1) sin2x = (-13 - 11) / 4 = - 24/4 = - 6 < - 1 - нет решений; 2) sin2x = (-13 + 11) / 4 = - 2/4 = - 1/2; 2x = - π/2 ± π/3 + 2πk, k ∈ Z; x = - π/4 ± π/6 + πk, k ∈ Z.

3)

sin^2 (2x - π/6) = 3/4; sin (2x - π/6) = ±√3/2; 2x - π/6 = ±π/3 + πk, k ∈ Z; 2x = π/6 ± π/3 + πk, k ∈ Z; x = π/12 ± π/6 + πk/2, k ∈ Z.

4)

sin^2 (x/2) = cos^2 (7x/2); 2cos^2 (7x/2) - 2sin^2 (x/2) = 0; (2cos^2 (7x/2) - 1) + (1 - 2sin^2 (x/2)) = 0; cos7x + cosx = 0; 2cos4x * cos3x = 0; [cos4x = 0;

[cos3x = 0; [4x = π/2 + πk, k ∈ Z;

[3x = π/2 + πk, k ∈ Z; [x = π/8 + πk/4, k ∈ Z;

[x = π/6 + πk/3, k ∈ Z.