Answer
$$c = \pi $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \cos x;{\text{ }}\left[ {\frac{\pi }{2},\frac{{3\pi }}{2}} \right],{\text{ }}a = \frac{\pi }{2},{\text{ }}b = \frac{{3\pi }}{2} \cr
& {\text{Calculating }}f'\left( x \right) \cr
& f'\left( x \right) = - \sin x \cr
& f'\left( c \right) = - \sin c \cr
& {\text{The function is continuous }} \cr
& f\left( a \right) = f\left( 0 \right) = \cos \left( {\frac{\pi }{2}} \right) = 0 \cr
& f\left( b \right) = f\left( 4 \right) = \cos \left( {\frac{{3\pi }}{2}} \right) = 0 \cr
& {\text{then there is at least one point }}c{\text{ in the interval}}\left( {a,b} \right){\text{ such that }} \cr
& f'\left( c \right) = 0 \cr
& - \sin c = 0 \cr
& c = \pi \cr
& {\text{Therefore, }} \cr
& f'\left( \pi \right) = 0 \cr} $$
