Provided (from the picture of a graph) that #f(0)=0# and #f((4pi)/3)=0#, we can write the following system of two equations with two unknown #x# and #c#:
#sin(0-k)+c = 0#
#sin((4pi)/3-k)+c = 0#
Subtracting one from another, we get a single equation with a single unknown #k#:
#sin(-k) - sin((4pi)/3-k) = 0#
This canbe simplified as
#-sin(k) - sin((4pi)/3)*cos(k)+cos((4pi)/3)*sin(k) = 0#
Since #sin((4pi)/3) = sin(pi/3+pi) = -sin(pi/3) = -sqrt(3)/2#
and #cos((4pi)/3) = cos(pi/3+pi) = -cos(pi/3) = -1/2#,
our equation looks like
#-sin(k)+sqrt(3)/2*cos(k)-1/2*sin(k) = 0#
Simplifying it to
#sqrt(3)/2*cos(k) = 3/2*sin(k)#
or
#sqrt(3)/3*cos(k) = sin(k)#
It's easy to prove that #cos(k)!=0# (otherwise, the second equation in our system would not be satisfied.)
So, dividing by #cos(k)#, we get
#tan(k) = sqrt(3)/3#
One of the solution of this equation is #k=pi/6#, then from the first equation of our system #c=sin(pi/6)=1/2#.
#f(x) = sin(x-pi/6)+1/2#
