1. cos3x – cos4x +cos5x = 0
pt <=> (cos 3x + cos5x) - cos 4x = 0
<=> 2cos4x.cosx - cos4x=0
<=> cos4x (2cosx-1)=0
=> cos4x= 0
or: 2cosx-1= 0
giải tiếp nhé.
2. cos3x . cos7x = sin4x . sin6x
pt <=> 1/2 (cos 10x + cos 4x ) = 1/2 (cos 10x - cos 2x )
<=> cos 4x = -cos 2x
<=> cos 4x = cos (pi-2x )
=> 4x= pi- 2x + k2pi
or 4x= 2x- pi+ k2pi
3. cos^2(x) + cos^2(2x) + cos^2(3x) +cos^2(4x) = 2
dùng công thức hạ bậc.
pt <=> $\dfrac{1+cos2x}{2} + \dfrac{1+cos4x}{2} + \dfrac{1+cos6x}{2} + \dfrac{1+cos8x}{2}= 2 $
<=> cos 2x + cos 4x + cos 6x cos 8x = 0
<=> (cos 2x + cos 8x) + (cos 4x+ cos 6x) =0
<=> 2 cos3x.cos5x + 2 cos 5x.cosx =0
<=>cos5x (cos 3x + cos x)=0
từ đó giải pt tích như câu 1