Answer
$$\eqalign{
& f\left( x \right){\text{ has a relative minimum at }}x = \frac{\pi }{4},\frac{{3\pi }}{4},\frac{{5\pi }}{4}{\text{,}}\frac{{7\pi }}{4} \cr
& f\left( x \right){\text{ has a relative maximum at }}x = x = \frac{\pi }{2},\pi ,\frac{{3\pi }}{2} \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\cos ^2}\left( {2x} \right) \cr
& \cr
& \left( {\text{a}} \right){\text{Calculating the first derivative}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{{\cos }^2}\left( {2x} \right)} \right] \cr
& f'\left( x \right) = 2\cos \left( {2x} \right)\left( { - \sin \left( {2x} \right)} \right) \cr
& f'\left( x \right) = - 2\sin \left( {2x} \right)\cos \left( {2x} \right) \cr
& {\text{Calculating the critical points, set }}f'\left( x \right) = 0 \cr
& f'\left( x \right) = \cos 2x \cr
& - 2\sin \left( {2x} \right)\cos \left( {2x} \right) = 0 \cr
& 2\sin \left( {2x} \right)\cos \left( {2x} \right) = 0 \cr
& {\text{On the interval }}\left( {0,2\pi } \right){\text{ }}\sin \left( {2x} \right) = 0,{\text{ for }}x = \frac{\pi }{2},\pi ,\frac{{3\pi }}{2} \cr
& {\text{On the interval }}\left( {0,2\pi } \right){\text{ }}\cos \left( {2x} \right) = 0,{\text{ for }}x = \frac{\pi }{4},\frac{{3\pi }}{4},\frac{{5\pi }}{4} \cr
& \cr
& \left( {\text{b}} \right){\text{Set the intervals: }} \cr
& \left( {0,\frac{\pi }{4}} \right),\left( {\frac{\pi }{4},\frac{\pi }{2}} \right),\left( {\frac{\pi }{2},\frac{{3\pi }}{4}} \right),\left( {\frac{{3\pi }}{4},\pi } \right),\left( {\pi ,\frac{{5\pi }}{4}} \right),\left( {\frac{{5\pi }}{4},\frac{{3\pi }}{2}} \right) \cr
& \left( {\frac{{3\pi }}{2},\frac{{7\pi }}{4}} \right),\left( {\frac{{7\pi }}{4},2\pi } \right) \cr} $$
\[\begin{array}{*{20}{c}}
{{\text{Interval}}}&{{\text{Test Value}}}&{{\text{Sign of }}f'\left( x \right)}&{{\text{Conclusion}}} \\
{\left( {0,\frac{\pi }{4}} \right)}&{x = \frac{\pi }{6}}&{ - \frac{{\sqrt 3 }}{2} < 0}&{{\text{Decreasing}}} \\
{\left( {\frac{\pi }{4},\frac{\pi }{2}} \right)}&{x = \frac{\pi }{3}}&{\frac{{\sqrt 3 }}{2} > 0}&{{\text{Increasing}}} \\
{\left( {\frac{\pi }{2},\frac{{3\pi }}{4}} \right)}&{x = \frac{{2\pi }}{3}}&{ - \frac{{\sqrt 3 }}{2} < 0}&{{\text{Decreasing}}} \\
{\left( {\frac{{3\pi }}{4},\pi } \right)}&{x = \frac{{5\pi }}{6}}&{\frac{{\sqrt 3 }}{2} > 0}&{{\text{Increasing}}} \\
{\left( {\pi ,\frac{{5\pi }}{4}} \right)}&{x = \frac{{7\pi }}{6}}&{ - \frac{{\sqrt 3 }}{2} < 0}&{{\text{Decreasing}}} \\
{\left( {\frac{{5\pi }}{4},\frac{{3\pi }}{2}} \right)}&{x = \frac{{4\pi }}{3}}&{\frac{{\sqrt 3 }}{2} > 0}&{{\text{Increasing}}} \\
{\left( {\frac{{3\pi }}{2},\frac{{7\pi }}{4}} \right)}&{x = \frac{{5\pi }}{3}}&{ - \frac{{\sqrt 3 }}{2} < 0}&{{\text{Decreasing}}} \\
{\left( {\frac{{7\pi }}{4},2\pi } \right)}&{x = \frac{{11\pi }}{6}}&{\frac{{\sqrt 3 }}{2} > 0}&{{\text{Increasing}}}
\end{array}\]
$$\eqalign{
& *f'\left( x \right){\text{ changes from negative to positive at }} \cr
& x = \frac{\pi }{4},\frac{{3\pi }}{4},\frac{{5\pi }}{4}{\text{,}}\frac{{7\pi }}{4}{\text{ }} \cr
& f\left( x \right){\text{ has a relative minimum at }}x = \frac{\pi }{4},\frac{{3\pi }}{4},\frac{{5\pi }}{4}{\text{,}}\frac{{7\pi }}{4} \cr
& *f'\left( x \right){\text{ changes from positive to negative at }} \cr
& x = \frac{\pi }{2},\pi ,\frac{{3\pi }}{2}{\text{}} \cr
& f\left( x \right){\text{ has a relative maximum at }}x = \frac{\pi }{2},\pi ,\frac{{3\pi }}{2} \cr
& \cr
& \left( {\text{c}} \right){\text{ Graph}} \cr} $$
