3, cos2x - 2√3sinxcosx = 2sin3x
<=>cos2x - √3sin2x = 2sin3x
<=>1/2.cos2x - √3/2sin2x = sin3x
<=>sin(pi/6-2x) =sin3x
3x =pi/6-2x +k2pi
5x =pi/6 +k2pi
x =pi/30 +k2pi/5
3x =pi -pi/6+2x +k2pi
x =5p /6 +k2pi
4, 2sin²x + √3sin2x = 3
<=>2sin²x + 2√3sin xcosx = 3
cosx =0 ko phải nghiệm
<=>2sin²x/cos^2x + 2√3sin xcosx /cos^2x= 3/cos^2x
<=>2tan²x + 2√3tanx= 3(tan^2 x +1)
<=> tan²x - 2√3tanx+3
<=> (tan x - √3)^2 =0
tanx =√3
x =pi/3 +kpi

Câu 1: le huy sai

Câu 4:
2sin²x + √3.sin2x = 3
<=> 2sin²x - 1 + √3sin2x = 2
<=> -(1 - 2sin²x) + √3.sin2x = 2
<=> -cos2x + √3.sin2x = 2
<=> √3.sin2x - cos2x = 2
Chia pt trên cho 2
<=> √3/2.sin2x - 1/2.cos2x = 1
Biến đổi : cosα = √3/2 và sinα = 1/2 => α = π/6
<=> cosπ/6.sin2x - sinπ/6.cos2x = 1
<=> sin( 2x - π/6 ) = 1
<=> 2x - π/6 = π/2 + k2π ( k thuộc Z )
<=> 2x = 2π/3 + k2π
<=> x = π/3 + kπ

Câu 5:
4(sin^4 x + cos^4 x) + √3sin4x = 2
<=> 4[(sin^2 x + cos^2 x)^2 - 2sin^2 x . cos^2 x] + √3sin4x = 2
<=> 4[1 - (1/2)sin^2 2x] + √3sin4x = 2
<=> -2sin^2 2x + √3sin4x = -2
<=> cos4x - 1 + √3sin4x = 2
<=> cos4x + √3sin4x = -1
<=> cos4x + tanpi/3.sin4x = -1
<=> cos4x.cospi/3 + sinpi/3.sin4x = -cospi/3
<=> cos(4x - pi/3) = cos(pi - pi/3) = cos(2pi/3)
=>
[4x - pi/3 = 2pi/3 + k2pi
[4x - pi/3 = -2pi/3 + k2pi
<=>
[x = pi/4 + kpi/2
[x = -pi/12 + kpi/2
với k thuộc Z