cos 3x= -cos (x+pi/4)

`cos3x=-cos(x+pi/4)`

`<=>` `cos3x=cos(x+pi/4+pi)`

`<=>` `cos3x=cos(x+(5pi)/4)`

TH1: `3x=x+(5pi)/4+k2pi` `(k in ZZ)`

`<=>` `2x=(5pi)/4+k2pi` `(k in ZZ)`

`<=>` `x=(5pi)/8+kpi` `(k in ZZ)`

TH2: `3x=-x-(5pi)/4+k2pi` `(k in ZZ)`

`<=>` `4x=-(5pi)/4+k2pi` `(k in ZZ)`

`<=>` `x=-(5pi)/16+(kpi)/2` `(k in ZZ)`

`<=>` `x=(3pi)/16+(kpi)/2` `(k in ZZ)`

Vậy ..............................................