1) 2sin^2x-5cosx-1=0 2) sin2x-√3cos2x=0 3) 3cos^2x-sin^2x+4sinx*cosx=0

1)

2sin^2x - 5cosx - 1 = 0; 2 - 2cos^2x - 5cosx - 1 = 0; 2cos^2x + 5cosx - 1 = 0; cosx = (-5 ± √ (5^2 + 4 * 2)) / 4 = (-5 ± √33) / 4; a) cosx = (-5 - √33) / 4 < - 1 - нет решения; b) cosx = (-5 + √33) / 4 = (√33 - 5) / 4; x = ±arccos ((√33 - 5) / 4) + 2πk, k ∈ Z.

2)

sin2x - √3cos2x = 0; sin2x = √3cos2x; tg2x = √3; 2x = π/3 + πk, k ∈ Z; x = π/6 + πk/2, k ∈ Z.

3)

3cos^2x - sin^2x + 4sinx * cosx = 0; 3 (1 + cos2x) / 2 - (1 - cos2x) / 2 + 4sinx * cosx = 0; 1 + 2cos2x + 2sin2x = 0; 2 (cos2x + sin2x) = - 1; cos2x + sin2x = - 1/2; √2/2 * (cos2x + sin2x) = - 1/2 * √2/2; sin (2x + π/4) = - √2/4; [2x + π/4 = - arcsin (√2/4) + 2πk, k ∈ Z;

[2x + π/4 = π + arcsin (√2/4) + 2πk, k ∈ Z; [2x = - π/4 - arcsin (√2/4) + 2πk, k ∈ Z;

[2x = - π/4 + π + arcsin (√2/4) + 2πk, k ∈ Z; [x = - π/8 - 1/2arcsin (√2/4) + πk, k ∈ Z;

[x = 3π/8 + 1/2arcsin (√2/4) + πk, k ∈ Z.