Answer
$$\eqalign{
& {\text{Maximum at: }}\left( {\frac{\pi }{4},1} \right){\text{ and }}\left( {\frac{{5\pi }}{4},1} \right) \cr
& {\text{Minimum at: }}\left( {\frac{{3\pi }}{4}, - 1} \right){\text{ and }}\left( {\frac{{7\pi }}{4}, - 1} \right) \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \sin 2x,{\text{ }}\left[ {0,2\pi } \right] \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\sin 2x} \right] \cr
& f'\left( x \right) = 2\cos 2x \cr
& {\text{Let }}f'\left( x \right) = 0 \cr
& 2\cos 2x = 0 \cr
& \cos 2x = 0 \cr
& {\text{For the interval }}\left[ {0,2\pi } \right]{\text{ }}\cos 2x = 0{\text{ when:}} \cr
& x = \frac{\pi }{4},\frac{{3\pi }}{4},\frac{{5\pi }}{4},\frac{{7\pi }}{4}{\text{ }}\left( {{\text{critical values}}} \right) \cr
& {\text{Evaluate }}f\left( x \right) = \sin 2x,{\text{ at the endpoints and critical}}{\text{ values}} \cr
& f\left( 0 \right) = \sin 2\left( 0 \right) = 0 \to \left( {0,0} \right) \cr
& f\left( {\frac{\pi }{4}} \right) = \sin 2\left( {\frac{\pi }{4}} \right) = 1 \to \left( {\frac{\pi }{4},1} \right) \cr
& f\left( {\frac{{3\pi }}{4}} \right) = \sin 2\left( {\frac{{3\pi }}{4}} \right) = - 1 \to \left( {\frac{{3\pi }}{4}, - 1} \right) \cr
& f\left( {\frac{{5\pi }}{4}} \right) = \sin 2\left( {\frac{{5\pi }}{4}} \right) = 1 \to \left( {\frac{{5\pi }}{4},1} \right) \cr
& f\left( {\frac{{7\pi }}{4}} \right) = \sin 2\left( {\frac{{7\pi }}{4}} \right) = - 1 \to \left( {\frac{{7\pi }}{4}, - 1} \right) \cr
& f\left( {2\pi } \right) = \sin 2\left( {2\pi } \right) = 0 \to \left( {2\pi ,0} \right) \cr
& {\text{Therefore}} \cr
& {\text{Maximum at: }}\left( {\frac{\pi }{4},1} \right){\text{ and }}\left( {\frac{{5\pi }}{4},1} \right) \cr
& {\text{Minimum at: }}\left( {\frac{{3\pi }}{4}, - 1} \right){\text{ and }}\left( {\frac{{7\pi }}{4}, - 1} \right) \cr} $$
