The graph is below. The curve with its "ripples" is shown in red. You can see the asymptotes at x=0 (y-axis) and x=π (vertical purple line) and periodically (period=π) before and after along the x-axis. I've also inserted the lines x=2 (blue), when y=cot(2)=-0.4577, and x=π/2 (vertical green).
dy/dx=2-cosec2(x)=0 at extrema. So sin2(x)=½, sin(x)=√2/2, x=π/4, 3π/4, 5π/4, ... at max or min.
y=2(x-2)-cot(x), cot(π/4)=1 cot(-π/4)=-1, y=2(x-2)±1=2x-5 or 2x-3. π/4=0.7854 approx. All the minima lie on the line y=2x-3 (green sloping line) and all the maxima on y=2x-5 (purple sloping line), so sketching these lines on the graph helps to draw the graph of the curve.
I think that whoever examines the answer to your problem would expect you to employ these facts of algebra, trigonometry and calculus to draw the graph with some degree of accuracy: asymptotes; max and min; periodicity.