The solutions of the equation `2cos(2x-(pi)/(3))+1=0` are
- `x=(pi(6k-2))/(6)" or "x=(pi(6k-3))/(6)," for "k in Z`
- `x=(pi(6k-2))/(6)" or "x=(pi(6k+5))/(6)," for "k in Z`
- `x=(pi(6k-1))/(6)" or "x=(pi(6k+2))/(6)," for "k in Z`
- `x=(pi(6k-1))/(6)" or "x=(pi(6k+3))/(6)," for "k in Z`
- `x=pi" or "x=(pi(6k+2))/(6)," for "k in Z`