We have cosx+cos2x+cos3x=0 or (cos3x+cosx)+cos2x=0 or 2cos2x⋅cosx+cos2x=0 or cos2x(2cosx+1)=0 We have either cos2x=0or2cosx+1=0 If
cos2x=0,then2x=(2m+1)
π
2
or x=(2m+1)
π
4
,m∈I If 2cosx+1=0,then cosx=−
1
2
=cos
2π
3
⇒x=2nπ±
2π
3
,n∈I Hence the required general solution are x=(2m+1)