Answer
$$\left( {\frac{\pi }{2} + n\pi ,0} \right),{\text{n is an integer}}$$
Work Step by Step
$$\eqalign{
& y = \cos x \cr
& {\text{Calculate the curvature, use }}K = \frac{{\left| {y''\left( x \right)} \right|}}{{{{\left( {1 + {{\left[ {y'\left( x \right)} \right]}^2}} \right)}^{3/2}}}} \cr
& {\text{Find }}y'\left( x \right){\text{ and }}y''\left( x \right),{\text{ }} \cr
& y'\left( x \right) = \frac{d}{{dx}}\left[ {\cos x} \right] \cr
& y'\left( x \right) = - \sin x \cr
& y''\left( x \right) = \frac{d}{{dx}}\left[ { - \sin x} \right] \cr
& y''\left( x \right) = - \cos x \cr
& \underbrace {K = \frac{{\left| {y''\left( x \right)} \right|}}{{{{\left( {1 + {{\left[ {y'\left( x \right)} \right]}^2}} \right)}^{3/2}}}}}_ \Downarrow \cr
& K = \frac{{\left| { - \cos x} \right|}}{{{{\left( {1 + {{\left[ { - \sin x} \right]}^2}} \right)}^{3/2}}}} \cr
& K = \frac{{\left| { - \cos x} \right|}}{{{{\left( {1 + {{\sin }^2}x} \right)}^{3/2}}}} \cr
& {\text{Find the point such that the curvature is zero}}{\text{.}} \cr
& 0 = \frac{{\left| { - \cos x} \right|}}{{{{\left( {1 + {{\sin }^2}x} \right)}^{3/2}}}} \cr
& \left| { - \cos x} \right| = 0 \cr
& {\text{Solving }}x = \frac{\pi }{2} + n\pi \cr
& y\left( {\frac{\pi }{2} + n\pi } \right) = \cos \left( {\frac{\pi }{2} + n\pi } \right) \cr
& y\left( {\frac{\pi }{2} + n\pi } \right) = 0 \cr
& {\text{The curvature is zero at the point }}\left( {\frac{\pi }{2} + n\pi ,0} \right) \cr} $$
