c) 3sinx +4cosx=5
x/2=t
<=>3.2.sint.cost +4( 2cos^2t-1) =5
6sint.cost +8cos^2t-9=0
cost =0 ko phai nghiẹm
chia cho cos^2t
<=>6tant +8 -9tan^2 t -9 =0
9tan^2 t -6tant +1 =0
(3tan t-1)^2 =0
tan t=1/3
t=artan[1/3 ]+kpi
x=2artan[1/3 ]+2kpi

3a) (2cosx-1)(2sinx+cosx) = sin2x - sinx
<=> 4cosxsinx + 2cos^2 x - 2sinx - cosx - sin2x + sinx = 0
<=> 2sin2x - sin2x - sinx + 2cos^2 x - cosx = 0
<=> sin2x - sinx + 2cos^2 x - cosx = 0
<=> 2sinxcosx - sinx + cosx(2cosx - 1) = 0
<=> sinx(2cosx - 1) + cosx(2cosx - 1) =0
<=> (2cosx - 1)(sinx + cosx) = 0
<=> 2cosx - 1 = 0        (1)
hoac sinx + cosx = 0    (2)
Giai? (1): 2cosx - 1 =0 <=> cosx = 1/2 <=> x = + - (pi/3 + k2pi)
Giai? (2): sinx + cosx = 0 <=> sinx = -cosx <=> tanx = -1 <=> x = - pi/4 + kpi

3b) cosx + cos2x +cos3x + cos4x=0
cosx + cos3x +cos2x+cos4x=0
2cos2xcosx + 2cos3xcosx=0
cosx(cos2x+cos3x)=0
cosx=0 hoặc cos2x+cos3x=0
x = pi/2 +kpi
hoặc cos2x= -cos3x
cos2x= cos(pi-3x)
x= pi/5 + k2pi/5 hoặc x=pi+ k2pi
vậy họ nghiệm là x= pi/2 + kpi , x= pi/5 + k2pi/5 , x= pi +k2pi

3c) 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