3sin4x+2cos23x+cos3x=3cos4x−cosx+1
⇔3(sin4x−cos4x)+2cos23x+cos3x+cosx−1=0
⇔3(sin2x−cos2x)+2cos23x+cos3x+cosx−1=0
⇔cos6x−3cos2x+cos3x+cosx=0
⇔(cos6x+cos2x)+(cos3x+cosx)−4cos2x=0
⇔2cos4xcos2x+2cos2xcosx−4cos2x=0
⇔cos2x(cos4x+cosx−2)=0
⇔[cos2x=0{cos4x=1cosx=1
⇔[cos2x=0cosx=1 dễ rồi tự làm