Answer
$$K = 4,{\text{ }}r = \frac{1}{4}$$
Work Step by Step
$$\eqalign{
& y = \cos 2x,{\text{ }}x = 2\pi \cr
& {\text{Calculate the curvature, use }}K = \frac{{\left| {y''\left( x \right)} \right|}}{{{{\left( {1 + {{\left[ {y'\left( x \right)} \right]}^2}} \right)}^{3/2}}}} \cr
& {\text{Find }}y'\left( x \right){\text{ and }}y''\left( x \right),{\text{ }} \cr
& y'\left( x \right) = \frac{d}{{dx}}\left[ {\cos 2x} \right] \cr
& y'\left( x \right) = - 2\sin 2x \cr
& {\text{Evaluate at }}x = 2\pi \cr
& y'\left( {2\pi } \right) = 0 \cr
& y''\left( x \right) = \frac{d}{{dx}}\left[ { - 2\sin 2x} \right] \cr
& y''\left( x \right) = 4\cos 2x \cr
& {\text{Evaluate at }}x = 2\pi \cr
& y''\left( {2\pi } \right) = 4\cos 2x \cr
& y''\left( {2\pi } \right) = 4\cos 4\pi = 4 \cr
& \underbrace {K = \frac{{\left| {y''\left( x \right)} \right|}}{{{{\left( {1 + {{\left[ {y'\left( x \right)} \right]}^2}} \right)}^{3/2}}}}}_ \Downarrow \cr
& {\text{ at }}x = 0 \cr
& K = \frac{{\left| 4 \right|}}{{{{\left( {1 + {{\left[ 0 \right]}^2}} \right)}^{3/2}}}} \cr
& K = 4 \cr
& {\text{The radius of curvature is }}r = \frac{1}{K} \cr
& r = \frac{1}{4} \cr} $$
