Toán Giải phưong trình: cosx + cos3x + 2cos5x = 0 13/07/2021 By Madeline Giải phưong trình: cosx + cos3x + 2cos5x = 0
`osx+cos3x+2cos5x=0cosx+cos3x+2cos5x=0` `cosx+cos3x+cos5x+cos5x=0cosx+cos3x+cos5x+cos5x=0` `cos3xcos2x+2cos4xcosx=02cos3xcos2x+2cos4xcosx=0` `cosx(4cos2x−3)cos2x+cos4xcosx=0cosx(4cos2x-3)cos2x+cos4xcosx=0` `cosx[(4cos2x−3)cos2x+cos4x]=0cosx[(4cos2x-3)cos2x+cos4x]=0` `cosx=0cosx=0` (`4cos2x−3)cos2x+cos4x=0(4cos2x-3)cos2x+cos4x=0` [`2(cos2x+1)−3]cos2x+2cos22x−1=0[2(cos2x+1)-3]cos2x+2cos22x-1=0` `2cos22x−cos2x+2cos22x−1=02cos22x-cos2x+2cos22x-1=0` `4cos22x−cos2x−1=04cos22x-cos2x-1=0` xin hay nhất ạ ! Trả lời
`osx+cos3x+2cos5x=0cosx+cos3x+2cos5x=0` `cosx+cos3x+cos5x+cos5x=0cosx+cos3x+cos5x+cos5x=0` `cos3xcos2x+2cos4xcosx=02cos3xcos2x+2cos4xcosx=0` `cosx(4cos2x−3)cos2x+cos4xcosx=0cosx(4cos2x-3)cos2x+cos4xcosx=0` `cosx[(4cos2x−3)cos2x+cos4x]=0cosx[(4cos2x-3)cos2x+cos4x]=0` `cosx=0cosx=0` (`4cos2x−3)cos2x+cos4x=0(4cos2x-3)cos2x+cos4x=0` [`2(cos2x+1)−3]cos2x+2cos22x−1=0[2(cos2x+1)-3]cos2x+2cos22x-1=0` `2cos22x−cos2x+2cos22x−1=02cos22x-cos2x+2cos22x-1=0` `4cos22x−cos2x−1=04cos22x-cos2x-1=0` xin hay nhất ạ !