Bài 1

2) sin 5x + sin 3x = sin 4x
<=> 2 . sin 4x . cos x - sin 4x = 0
<=> sin 4x (2 cos x - 1) = 0
<=> sin 4x = 0 or 2 cos x = 1
<=> 4x = kπ or x = ±π/3 + k2π
<=> x = kπ/4 or x = ±π/3 + k2π
3) Trên ảnh

7) 1 + cosx + cos2x + cos3x = 0
<=> ( cosx + cos3x) + 1 + cos2x = 0
<=> 2cos(x + 3x)/2.cos(x - 3x)/2 + 1 + cos2x = 0
<=> 2cos2x.cosx + 1 + cos2x = 0
<=> 2cos2x.cosx + 2cos^2x = 0 
<=> 2cos2x(cosx + cos2x) = 0
<=> 2cos2x.( 2cos(x +2x)/2.cos(x - 2x)/2 ) = 0
<=> 2cos2x.2cos3x/2.cosx/2 = 0
<=> 2cos2x = 0
Hoac 2cos3x/2 = 0
      cosx/2 = 0
<=> cos2x = 0
Hoac cos3x/2 = 0
        cosx/2 = 0
<=> 2x = π/2 + kπ
Hoac 3x/2 = π/2 + kπ ( k thuoc Z )
          x/2 = π/2 + kπ
<=> x = π/4 + kπ/2
Hoac x = π/3 + 2kπ/3 ( k thuoc Z )
         x = π + 2kπ
4) Trên ảnh

8) Trên ảnh
9) 1 + sinx + cosx + sin2x + cos2x / tan2x = 0
<=>1 + sinx + cosx + sin2x + cos^2 2x / sin2x = 0
<=>1 + sinx + cosx + 1/ sin2x = 0
(sinx+cosx)=t  |t |≤√2
sin2x=t^2-1
<=>1+t+1/(t^2-1)=0
t≠±1
(t^2-1)(t+1)+1=0
t^3+t^2-t-1+1=t(t^2+t-1)=0
t=0; sin x +cos x=0; x=-π/4+kπ
t=(-1-√5)/2;(loai)
t=(√5-1)/2 (nhan)
sinx+cosx =(√5-1)/2 
sin(x+π/4)==(√10-√2)/4 
x=-π /4+acrsin((√10-√2)/4 )+kπ

12) tanx + tan2x = tan 3x
<=>sinx/cosx + sin2x/cos2x = sin3x/cos3x
<=>sin3x/cosxcos2x = sin3x/cos3x
<=>sin3x(cosxcos2x - cos3x) = 0
<=>sin3x(sinxsin2x) = 0
<=>sinx=0 hoặc sin2x=0 hoặc sin3x=0
<=> x = kpi or x = kpi/2 or x = kpi/3 (k thuộc Z)
14) Trên ảnh

15) sinx + sin2x + sin3x = cosx + cos2x + cos3x
<=>sin3x + sinx + sin2x = cos3x + cosx + cos2x
<=>2sin2xcosx + sin2x = 2cos2xcosx + cos2x
<=>sinx(2cosx + 1) = cosx(2cosx + 1)
<=>(sinx - cosx)(2cosx + 1) = 0
<=> sinx-cosx=0 or 2cosx+1=0
<=> x = pi/4 + kpi or x = 2pi/3 + 2kpi or x = 4pi/3 + 2kpi (k thuộc Z)
16) cos10x - cos8x = cos6x - 1
<=> cos10x - cos8x - cos6x + 1 = 0
<=> cos 10x - cos 6x + 1 - cos 8x = 0
<=> -2.sin 8x.sin 2x + 2.sin^2(4x) = 0
<=> -2.sin4x.cos 4x.sin 2x + sin^2(4x) = 0
<=> sin 4x( sin 4x- 2cos 4x.sin 2x) = 0
<=> sin 4x ( 2.sin 2x . cos 2x - 2.cos 4x .sin 2x) = 0
<=> sin 4x .sin 2x. (cos 2x - cos 4x) = 0
<=> sin 4x . sin 2x . ( - 2.cos^2(2x) + cos 2x + 1) = 0
<=> sin 4x = 0 hoặc sin 2x = 0 hoặc cos 2x = 1 hoặc cos 2x = -1/2

1) va 3) Trên ảnh
2) sinx+sin2x+sin3x+sin4x+sin5x+sin6x=0
<=> sinx +sin6x + sin2x +sin5x + sin3x+sin4x = 0
<=> 2sin(7x/2).cos(5x/2) + 2sin(7x/2).cos(3x/2) + 2.sin(7x/2).cos(x/2) = 0
<=> 2sin(7x/2)[cos(5x/2) + cos(3x/2) + cos(x/2) = 0
<=> 2sin(7x/2)[2cos(2x).cos(x/2) + cos(x/2)] = 0
<=> 2sin(7x/2).cos(x/2)[2cos(2x) + 1] = 0
<=>
{ sin(7x/2) = 0 => 7x/2 = kπ => x = k2π/7
{ cos(x/2) = 0 => x/2 = (2k+1)π/2 => x = (2k+1)π
{2cos(2x) + 1 <=> cos(2x) = -1/2 = cos(2π/3) => x = ± π/3 + k2π