1. Sin (3x + 1) = sin(x - 2)
=> 3x + 1 = x - 2 + k2π
<=> 2x = -3 + k2π
<=> x = -3/2 + kπ (k∈Z)
và 3x + 1 = π - (x - 2) + k2π
<=> 3x + 1 = π - x + 2 + k2π
<=> 4x = π + 1 + k2π
<=> x = π/4 + 1/4 + kπ/2 (k∈Z)
vậy phương trình có nghiệm là
x = -3/2 + kπ (k∈Z)
x = π/4 + 1/4 + kπ/2 (k∈Z)

2. Cos (x - π/3) = cos (2x + π/6)
=> x - π/3 = 2x + π/6 + k2π
<=> -x = π/2 + k2π
<=> x = - π/2 - k2π (k ∈ Z)
x - π/3 = -(2x + π/6) + k2π
<=> x - π/3 = -2x - π/6 + k2π
<=> 3x = π/6 + k2π
<=> x = π/18 + k2π/3 (k ∈ Z)
vậy phương trình có nghiệm là
x = - π/2 - k2π (k ∈ Z)
x = π/18 + k2π/3 (k ∈ Z)

3)
cos3x = sin2x
 <=> cos3x = cos(π/2-2x) (hai góc phụ nhau)
<=> [3x = π/2 - 2x + k2π (k ∈ Z)
        [3x = -π/2 + 2x + k2π
<=> [x = π/10 + kπ/5 
        [x = -π/2 + k2π
Vậy...

4)
sin(x - 120°) + cos2x = 0
=> sin(x-120°) = -cos2x
=> sin(x - 120°) = cos(180° - 2x)
=> sin(x - 120°) = sin(90° - 180° + 2x)
=> sin(x - 120°) = sin(2x - 90°)
=> x - 120° = 2x - 90° + k360°
   hoặc x - 120° = 180° - 2x + 90° + k360°
=> x = -30° + k360° hoặc x =  130° + k120°  (k € Z)
Vậy . . .
 

6)
sin3x + sin(π/4 + x/2) = 0
<=>sin3x = -sin (π/4 - x/2)
<=> sin3x = sin(x/2 - π/4)
<=> [3x = x/2- π/4 + k2π (k ∈ Z)
       [3x= 5π/4 - x/2 + k2π
<=> [x= -π/10 + 4kπ/5
       [x= 5π/14 + 4kπ/7 
Vậy...

7)
tan(2x + π/4) + cotx = 0 (1)
Đk: +) cos ( 2x + π/4) # 0
<=> 2x + π/4 # (π/2) + kπ
<=> x # π/8 + kπ/2     (k € Z)
+) sin x # 0 <=> x # kπ
PT (1) <=> (sin(2x + π/4)/cos(2x + π/4)) + cosx/sinx = 0 
<=> sin(2x +π/4 ).sinx + cos(2x +  π/4).cosx = 0 
<=> cos (2x +  π/4 - x) = 0 
<=> cos (x + π/4) = 0
<=> x + π/4 = (π/2) + kπ 
<=> x = π/4+ kπ (k ∈ Z) ( TMĐK)

8)
sin^2(x-pi/4) = cos^2x
=> sin^2(x-pi/4)-cos^2x=0
=> [sin(x-pi/4)+cosx][sin(x-pi/4)-cosx]=0
=> sin(x-pi/4)+cosx=0 (*)
     hoặc sin(x-pi/4)-cosx=0 (**)
+) từ (*) => sin(x-pi/4)=-cosx
=>sin(x-pi/4)=cos(pi-x)
=> sin(x-pi/4)=sin(pi/2-pi+x)
=> sin(x-pi/4)=sin(x-pi/2) (vô lí =>loại)
+) từ (**) => sin(x-pi/4)=cosx
=> sin(x-pi/4)=sin(pi/2-x)
=>...tự giải tiếp nha...

6)
sin3x + sin(π/4 + x/2) = 0
<=> sin(3x) = -sin(pi/4 - x/2)
<=> sin(3x) = sin(x/2 - pi/4)
<=> 3x = x/2 - pi/4 + k2pi hoặc 3x = 5pi/4 - x/2 + k2pi
từ đó ta thu được các nghiệm của phương trình là:
x = -pi/10 + 4kpi/5 và x = 5pi/14 + 4kpi/7

3)
cos3x = sin2x
<=> cos(π/2 - 2x) " vì hai góc phụ nhau" 
<=> [3x = π/2 - 2x + k2π   <=>    [x = π/10 + π/5 
       [3x =-π/2 + 2x + k2π             [x = -π/2 + k2π
Vậy.....

6)sin3x + sin(π/4 + x/2) = 0
<=>sin3x = -sin (π/4 - x/2)
<=> sin3x = sin(x/2 - π/4)
<=> [3x = x/2- π/4 + k2π (k ∈ Z)
       [3x= 5π/4 - x/2 + k2π
<=> [x= -π/10 + 4kπ/5
       [x= 5π/14 + 4kπ/7 
Vậy...
7)
tan(2x + π/4) + cotx = 0 (1)
Đk: +) cos ( 2x + π/4) # 0
<=> 2x + π/4 # (π/2) + kπ
<=> x # π/8 + kπ/2     (k € Z)
+) sin x # 0 <=> x # kπ
PT (1) <=> (sin(2x + π/4)/cos(2x + π/4)) + cosx/sinx = 0 
<=> sin(2x +π/4 ).sinx + cos(2x +  π/4).cosx = 0 
<=> cos (2x +  π/4 - x) = 0 
<=> cos (x + π/4) = 0
<=> x + π/4 = (π/2) + kπ 
<=> x = π/4+ kπ (k ∈ Z) ( TMĐK)