Phương pháp giải:

\[\sin \left( {a \pm b} \right) = \sin a\cos b \pm \cos a\sin b\]

\[\cos \left( {a \pm b} \right) = \cos a\cos b \mp \sin a\sin b\]

Giải chi tiết:

\[y = \cos 2x + \sin {\mkern 1mu} x - \sqrt 3 \left( {\sin 2x + \cos x} \right) + 3\]\[ \Leftrightarrow y = \left( {\cos 2x - \sqrt 3 \sin 2x} \right) + \left( {\sin {\mkern 1mu} x - \sqrt 3 \cos x} \right) + 3\]

\[ \Leftrightarrow y = - 2\left( {\frac{{ - 1}}{2}\cos 2x + \frac{{\sqrt 3 }}{2}\sin 2x} \right) + 2\left( {\frac{1}{2}\sin {\mkern 1mu} x - \frac{{\sqrt 3 }}{2}\cos x} \right) + 3\]

\[ \Leftrightarrow y = - 2\left( {\cos \frac{{2\pi }}{3}\cos 2x + \sin \frac{{2\pi }}{3}\sin 2x} \right) + 2\left( {\cos \frac{\pi }{3}\sin {\mkern 1mu} x - \sin \frac{\pi }{3}\cos x} \right) + 3\]

\[ \Leftrightarrow y = - 2\cos \left( {2x - \frac{{2\pi }}{3}} \right) + 2\sin \left( {x - \frac{\pi }{3}} \right) + 3\]\[ \Leftrightarrow y = 4{\sin ^2}\left( {x - \frac{\pi }{3}} \right) + 2\sin \left( {x - \frac{\pi }{3}} \right) + 1\]

Ta có: \[x \in \left[ { - \frac{\pi }{3};\frac{\pi }{2}} \right] \Leftrightarrow x - \frac{\pi }{3} \in \left[ { - \frac{{2\pi }}{3};\frac{\pi }{6}} \right]{\mkern 1mu} \]

Khi đó: \[\sin \left( {x - \frac{\pi }{3}} \right) \in \left[ { - \frac{{\sqrt 3 }}{2};\frac{1}{2}} \right]{\mkern 1mu} \Leftrightarrow 2\sin \left( {x - \frac{\pi }{3}} \right) \in \left[ { - \sqrt 3 ;1} \right]{\mkern 1mu} \]

Xét hàm số \[f\left( t \right) = {t^2} + t + 1,{\mkern 1mu} {\mkern 1mu} t \in \left[ { - \sqrt 3 ;1} \right],{\mkern 1mu} {\mkern 1mu} f'\left( t \right) = 2t + 1 = 0 \Leftrightarrow t = - \frac{1}{2}\]

Hàm số \[f\left( t \right)\] liên tục trên đoạn \[\left[ { - \sqrt 3 ;1} \right]\] có \[f\left( { - \sqrt 3 } \right) = 4 - \sqrt 3 ,{\mkern 1mu} {\mkern 1mu} f\left( { - \frac{1}{2}} \right) = \frac{3}{4},{\mkern 1mu} {\mkern 1mu} f\left( 1 \right) = 3\]

\[ \Rightarrow \mathop {\min }\limits_{\left[ { - \sqrt 3 ;1} \right]} f\left( t \right) = f\left( { - \frac{1}{2}} \right) = \frac{3}{4},{\mkern 1mu} {\mkern 1mu} \mathop {\max }\limits_{\left[ { - \sqrt 3 ;1} \right]} f\left( t \right) = f\left( 1 \right) = 3\]\[ \Rightarrow \mathop {\min }\limits_{\left[ { - \frac{\pi }{3};\frac{\pi }{2}} \right]} y = \frac{3}{4},{\mkern 1mu} {\mkern 1mu} \mathop {\max }\limits_{\left[ { - \frac{\pi }{3};\frac{\pi }{2}} \right]} y = 3\]