Giải phương trình cos2x + cos4x +cos6x =cosxcos2xcos3x + 2 A. x = kpi (k thuộc Z)

cos2x+cos4x+cos6x=cosxcos2xcos3x+2

⇔2cos4xcos2x+cos4x=12cos2xcos4x+cos2x+2

⇔2cos4xcos2x+cos4x=12cos2xcos4x+12cos22x+2

⇔32cos4xcos2x+cos4x=12cos22x+2

⇔3cos4xcos2x+2cos4x=cos22x+4

⇔32cos22x−1cos2x+22cos22x−1=cos22x+4

⇔6cos32x−3cos2x+4cos22x−2=cos22x+4

⇔6cos32x+3cos22x−3cos2x−6=0

⇔2cos32x+cos22x−cos2x−2=0

⇔2cos32x−1+cos2xcos2x−1=0

⇔2cos2x−1cos22x+cos2x+1+cos2xcos2x−1=0

⇔cos2x−12cos22x+2cos2x+2+cos2x=0

⇔cos2x−12cos22x+3cos2x+2=0

⇔cos2x=1

⇔2x=k2π

⇔x=kπk∈z

Vậy nghiệm của phương trình đã cho là: x = kπ (k ∈ Z).

Đáp án cần chọn là: A