3) 2cos²2x + 3cos2x + 1 = 0
<=> (2cos2x + 1)(cos2x + 1) = 0
<=> 2cos2x + 1 = 0
<=> Cos2x = -1/2
<=> Cos2x = cos2π/3
<=> 2x = ±2π/3 + k2π
<=> x = ±π/3 + kπ , k thuộc Z
Hoặc cos2x + 1 = 0
<=> cos2x = -1
<=> 2x = π + k2π
<=> x = π/2 + kπ , k thuộc Z
Vậy S = { -π/3 + kπ ; π/3 + kπ ; π/2 + kπ | k thuộc Z }

1. sin2x + sin^2x = 1/2
<=> sin2x = 1/2 - sin^2x
<=> sin2x = (cos2x)/2
<=> sin2x/cos2x = 1/2
<=> tan2x = 1/2
<=> tan2x = arctan1/2 + kπ
<=> x = (arctan1/2)/2 + kπ2, k nguyên
2. cos2x - 4cosx + 5/2 = 0
<=> 2cos^2x -1 - 4cosx + 5/2 = 0
<=> 2cos^2x - 4cosx + 3/2 = 0
<=> 1/2.(2cosx - 3)(2cosx - 1) = 0
<=> 2cosx - 1 = 0 ( do -5 < 2cosx - 3 < -1 với mọi x )
<=> cosx = 1/2
<=> cosx = cosπ/3
<=> x = ±π/3 + k2π, k nguyên

3. 2cos^2(2x) + 3cos2x + 1 = 0
<=> (cos2x + 1)(2cos2x + 1) = 0
<=> cos2x = -1 hoặc cos2x = -1/2 = cos2π/3
<=> 2x = π + k2π hoặc 2x = 2π/3 + k2π hoặc 2x = -2π/3 + k2π, k nguyên
<=> x = π/2 + kπ hoặc x = ±π/3 + kπ, k nguyên
4. sin^2x - 3sinx.cosx + 2cos^2x = 0
Ta thấy cosx = 0 không là nghiệm của pt nên chia cả 2 vế cho cos^2x, ta được
tan^2x - 3tanx + 2 = 0
<=> (tanx - 1)(tanx - 2) = 0
<=> tanx = 1 hoặc tanx = 2
<=> x = π/4 + kπ hoặc x = arctan2 + kπ, k nguyên

6. 2cos3x + √3.sinx + cosx = 0
<=> sinx.√3 + cosx = -2cos3x
<=> sinx.√3/2 + cosx.1/2 = -cos3x
<=> sinx.cosπ/3 + cosx.sinπ/3 = -cos3x
<=> sin(x + π/3) = -cos3x
<=> cos3x = -sin(x + π/3)
<=> cos3x = sin(-x - π/3)
<=> cos3x = cos(5π/6 + x)
<=> 3x = 5π/6 + x + k2π hoặc 3x = -5π/6 - x + k2π, k nguyên
<=> 2x = 5π/6 + k2π hoặc 4x = -5π/6 + k2π, k nguyên
<=> x = 5π/12 + kπ hoặc x = -5π/24 + kπ/2, k nguyên