3,
cos^2 x + cos^2 2x + cos^2 3x + cos^2 4x = 3/2 
<=> 1 + cos 2x + cos^2 2x + 1 + cos 6x + cos^2 4x = 3/2 
<=> cos 2x + cos^2 2x + cos 6x + (2cos^2 2x - 1)^2 = 1 
<=> cos 2x + cos^2 2x + (- 3cos 2x + 4 cos^3 2x) + (4cos^4 2x - 4cos^2 2x + 1) = 1 
<=> 4cos^4 2x + 4cos^3 2x - 3cos^2 2x - 2cos 2x = 0 
<=> cos 2x ( 4cos^3 2x+ 4cos^2 2x - 3cos 2x - 2 ) = 0 
=> ....cosx = 0 .......... 

4,
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

5, 
C1:
 4cosx -2cos2x-cos4x=1 
<=> 4cosx - 2[2cos^2(x) - 1] - [ 2cos^2(2x) - 1] = 1 
<=> 4cosx - 4cos^2(x) + 3 - 2cos^2(2x) = 1 
<=> 4cosx - 4cos^2(x)+ 2 - 2*[2cos^2(x) -1]^2 = 0 
<=> 4cosx - 4cos^2(x) + 2 - 2*(4cos^4(x) - 4cos^2 +1) = 0 
<=> 4cosx +4cos^2(x) - 8cos^4(x) = 0 
<=> cosx = 0 <=> x= pi/2 + k*pi 
hoặc cosx = 1 <=> x= k*pi
C2:
4cosx - 2(2(cosx)^2 -1) - 2(cos2x)^2 =0 
--> 4 cosx -4(cosx)^2 +2 -2 (2(cosx)^2 -1 )^2 =0 
đặt cosx=t ; t thuộc [-1;1] 
-> 4t -4t^2 +2 -2(4t^4-4t^2+1) =0 
-> 8t^4 -4t^2 -4t=0 --> 2t^4 -t^2-t =0 
-> t= 0 hoặc t =1 
--> cosx = 0 hoặc cosx=1 :D

6,
PT
<=>(cos^2x)^2-2cos^2x+1+2(1-cos^2x)^3 
đặt cos^2x=x.ta có:x^2-2x+1+2(1-x)^3=0 
<=>x^2-2x+1+2-6x+6x^2-2x^3=0 
<=>7x^2-2x^3-8x+3=0 
<=>x=3/2(loại) hoặc x=1( nhận) 
cos^2x=1
<=> 1/2(1+cos2x)=1
<=>cos2x=1 
<=>x=kpi

7,
Cos^2 (x) + cos ^2 (2x) + cos^2 (3x) + cos^2 (4x) = 2?
<=>1+cos 2x + 1+ cos 4x + 1+ cos 6x + 1 + cos 8x = 4 ( hạ bậc) 
<=> ( cos 2x + cos 8x) + ( cos 4x + cos 6x) = 0 
<=> 2cos 5x.cos 3x + 2cos 5x. cos x = 0 
<=> 2cos 5x. ( cos 3x + cos x) = 0 
<=> 4cos 5x. cos 2x. cos x = 0 
<=> cos 5x = 0 hoặc cos 2x = 0 hoặc cos x = 0

8,
sin²4x + sin²3x = sin²2x + sin²x 
Cách 1: hạ bậc 
pt ⇔ (1 - cos8x)/2 + (1 - cos6x)/2 = (1 - cos4x)/2 + (1 - cos2x)/2 
⇔ cos2x + cos4x = cos6x + cos8x 
⇔ 2cos3x.cosx = 2cos7x.cosx 
⇔ 2cosx.(cos3x - cos7x) = 0 
⇔ 2cosx.(-2).sin5x.sin(-2x) = 0 
⇔ cosx.sin2x.sin5x = 0 
⇔ sin2x.sin5x = 0, do sin2x = 0 ⇔ 2sinx.cosx = 0, đã bao gồm TH cosx = 0 rồi 
⇔ [ sin2x = 0 → x = kπ/2 (k ∈ Z) 
. . .[ sin5x = 0 → x = kπ/5 (k ∈ Z) 

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Cách 2: nhóm 
pt ⇔ (sin²4x - sin²2x) + (sin²3x - sin²x) = 0 
⇔ (sin4x - sin2x)(sin4x + sin2x) + (sin3x - sinx)(sin3x + sinx) = 0 
⇔ (2cos3x.sinx).(2sin3x.cosx) + (2cos2x.sinx).(2sin2x.cosx) = 0 
⇔ 2sinx.cosx.(2cos3x.sin3x + 2cos2x.sin2x) = 0 
⇔ sin2x.(sin6x + sin4x) = 0 
⇔ sin2x.2sin5x.cosx = 0 
⇔ sin2x.sin5x = 0 
Tương tự như trên 
Vậy pt đã cho có nghiệm là: x = kπ/2 (k ∈ Z) ; x = kπ/5 (k ∈ Z) 
 

9,
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

8.