Câu 2
c. 3sinx - 4cosx = 1
<=> sinx.3/5 - cosx.4/5 = 1/5
  Do (3/5)^2 + (4/5)^2 = 1 nên đặt cosa = 3/5 và sina = 4/5
  Phương trình trở thành
       sinx.cosa - cosx.sina = 1/5
<=> sin(x - a) = 1/5
<=> x - a = arcsin1/5 + k2π
hoặc x - a = π - arcsin1/5 + k2π, k nguyên
<=> x = arcsin1/5 + a + k2π hoặc x = π + a - arcsin1/5 + k2π, k nguyên với a = arccos3/5
d. √3.sin3x + cos3x = √2
<=> sin3x.√3/2 + cos3x.1/2 = √2/2
<=> sin3x.cosπ/6 + cos3x.sinπ/6 = sinπ/4
<=> sin(3x + π/6) = sinπ/4
<=> 3x + π/6 = π/4 + k2π
hoặc 3x + π/6 = 3π/4 + k2π, k nguyên
<=> x = π/36 + k2π/3 hoặc x = 7π/36 + k2π/3, k nguyên

Câu 3
a. cosx + √3.sinx = √3
<=> cosx.1/2 + sinx.√3/2 = √3/2
<=> cosx.cosπ/3 + sinx.sinπ/3 = cosπ/6
<=> cos(x - π/3) = cosπ/6
<=> x - π/3 = π/6 + k2π
hoặc x - π/3 = -π/6 + k2π, k nguyên
<=> x = π/2 + k2π hoặc x = π/6 + k2π, k nguyên
b. √3.sinx - cosx = √2
<=> sinx.√3/2 - cosx.1/2 = √2/2
<=> sinx.cosπ/6 - cosx.sinπ/6 = sinπ/4
<=> sin(x - π/6) = sinπ/4
<=> x - π/6 = π/4 + k2π 
hoặc x - π/6 = 3π/4 + k2π, k nguyên
<=> x = 5π/12 + k2π hoặc x = 11π/12 + k2π, k nguyên

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

e. √3.sinx + cosx = 2sin7x
<=> sinx.√3/2 + cosx.1/2 = sin7x
<=> sinx.cosπ/6 + cosx.sinπ/6 = sin7x
<=> sin(x + π/6) = sin7x
<=> 7x = x + π/6 + k2π
hoặc 7x = 5π/6 - x + k2π, k nguyên
<=> x = π/36 + kπ/3 hoặc x = 5π/48 + kπ/4, k nguyên
f. √2.cos13x = sinx + cosx
<=> cos13x = cosx.1/√2 + sinx.1/√2
<=> cos13x = cosx.cosπ/4 + sinx.sinπ/4
<=> cos13x = cos(x - π/4)
<=> 13x = x - π/4 + k2π
hoặc 13x = π/4 - x + k2π, k nguyên
<=> x = -π/48 + kπ/6 hoặc x = π/56 + kπ/7, k nguyên

g. √2.sin3x - √6.cos3x = -2
<=> sin3x.1/2 - cos3x.√3/2 = -√2/2
<=> sin3x.cosπ/3 - cos3x.sinπ/3 = sin(-π/4)
<=> sin(3x - π/3) = sin(-π/4)
<=> 3x - π/3 = -π/4 + k2π
hoặc 3x - π/3 = 5π/4 + k2π, k nguyên
<=> x = π/36 + k2π/3 hoặc x = 19π/36 + k2π/3, k nguyên
h. √2.sin4x - √2.cos4x = -1
<=> sin4x.1/√2 - cos4x.1/√2 = -1/2
<=> sin4x.cosπ/4 - cos4x.sinπ/4 = sin(-π/6)
<=> sin(4x - π/4) = sin(-π/6)
<=> 4x - π/4 = -π/6 + k2π
hoặc 4x - π/4 = 7π/6 + k2π, k nguyên
<=> x = π/48 + kπ/2 hoặc x = 17π/48 + kπ/2, k nguyên

Câu 4
a. sin3x - √3.cos3x = 2sin2x
<=> sin3x.1/2 - cos3x.√3/2 = sin2x
<=> sin3x.cosπ/3 - cos3x.sinπ/3 = sin2x
<=> sin(3x - π/3) = sin2x
<=> 3x - π/3 = 2x + k2π
hoặc 3x - π/3 = π - 2x + k2π, k nguyên
<=> x = π/3 + k2π hoặc x = 4π/15 + k2π/5, k nguyên
b. √3.sin3x - cos3x - √2 = 0
<=> sin3x.√3/2 - cos3x.1/2 = √2/2
<=> sin3x.cosπ/6 - cos3x.sinπ/6 = sinπ/4
<=> sin(3x - π/6) = sinπ/4
<=> 3x - π/6 = π/4 + k2π
hoặc 3x - π/6 = 3π/4 + k2π, k nguyên
<=> x = 5π/36 + k2π/3 hoặc x = 11π/36 + k2π/3, k nguyên

c. 2sin2x.cos2x + √3.cos4x = √2
<=> sin4x + √3.cos4x = √2
<=> sin4x.1/2 + cos4x.√3/2 = √2/2
<=> sin4x.cosπ/3 + cos4x.sinπ/3 = sinπ/4
<=> sin(4x + π/3) = sinπ/4
<=> 4x + π/3 = π/4 + k2π 
hoặc 4x + π/3 = 3π/4 + k2π, k nguyên
<=> x = -π/48 + kπ/2 hoặc x = 5π/48 + kπ/2, k nguyên
d. √3.cosx/2 - √2.sinx/2 = -√5
<=> cosx/2.√3/5 - sinx/2.√2/5 = -1
  Do (√3/5)^2 + (√2/5)^2 = 1 nên đặt cosa = √3/5 và sina = √2/5
  Phương trình trở thành
       cosx/2.cosa - sinx/2.sina = -1
<=> cos(x/2 + a) = -1
<=> x/2 + a = π + k2π, k nguyên
<=> x = 2π - 2a + k4π, k nguyên 

e. √3.cosx + sinx = 0
<=> cosx.√3 /2 + sinx.1/2 = 0
<=> cosx.cosπ/6 + sinx.sinπ/6 = 0
<=> cos(x - π/6) = 0
<=> x - π/6 = π/2 + kπ, k nguyên
<=> x = 2π/3 + kπ, k nguyên
f. sin4x + √3.cos4x = -√2
<=> sin4x.1/2 + cos4x.√3/2 = -√2/2
<=> sin4x.cosπ/3 + cos4x.sinπ/3 = sin(-π/4)
<=> sin(4x + π/3) = sin(-π/4)
<=> 4x + π/3 = -π/4 + k2π
hoặc 4x + π/3 = 5π/4 + k2π, k nguyên
<=> x = -7π/48 + kπ/2 hoặc x = 11π/48 + kπ/2, k nguyên

2
a) sinx + √3cosx = 1
<=> 1/2sinx + √3/2cosx = 1/2
<=> cosπ/3.sinx + sinπ/3.cosx = sinπ/6
<=> sin(x + π/3) = sinπ/6
<=> x + π/3 = π/6 + k2π
<=> x = -π/6 + k2π , k thuộc Z
hoặc x + π/3 = π - π/6 + k2π
<=> x  = π/2 + k2π , k thuộc Z
vậy S = { -π/6 + k2π ; π/2 + k2π | k thuộc Z }
d) √3sin3x + cos3x = √2
<=> √3/2sin3x + 1/2cos3x = √2
<=> cosπ/6.sin3x + sinπ/6.cos3x = sinπ/4
<=> sin(3x + π/6) = sinπ/4
<=> 3x + π/6 = π/4 + k2π <=> 3x = π/12 + k2π <=> x = π/36 + k2π/3 , k thuộc Z
hoặc 3x + π/6 = π - π/4 + k2π <=> 3x = 7π/12 + k2π <=> x = 7π/36 + k2π/3 , k thuộc Z
vậy ..