Câu 3:
sin2x - √3.cos2x = 1
<=> 1/2.sin2x - √3/2.cos2x = 1/2
<=> sin(2x - π/3) = sin(π/6)
<=> 2x - π/3 = π/6 + k2π hay 2x - π/3 = 5π/6 + k2π
<=> 2x = π/2 + k2π hay 2x = 7π/6 + k2π
<=> x = π/4 + kπ hay x = 7π/12 + kπ
Vậy..
Câu 5:
cos2x - 2√3.sinxcosx = 2sin3x
<=> cos2x - √3.sin2x = 2sin3x
<=> 1/2.cos2x - √3/2.sin2x = sin3x
<=> sin(π/6 - 2x) = sin3x
<=> π/6 - 2x = 3x + k2π hay π/6 - 2x = π - 3x + k2π
<=> -5x = -π/6 + k2π hay x = 5π/6 + k2π
<=> x = π/30 - k2π/5 hay x = 5π/6 + k2π
Vậy...