Tìm minmax..y=cos^2 (x+pi/3)+cos^2(x-pi/3)+cos^2(x-2pi/3)+3cos2x +1
1 câu trả lời
$$\eqalign{ & y = {\cos ^2}\left( {x + {\pi \over 3}} \right) + {\cos ^2}\left( {x - {\pi \over 3}} \right) + {\cos ^2}\left( {x - {{2\pi } \over 3}} \right) + 3\cos 2x + 1 \cr & y = {{1 + \cos \left( {2x + {{2\pi } \over 3}} \right) + 1 + \cos \left( {2x - {{2\pi } \over 3}} \right) + 1 + \cos \left( {2x - {{4\pi } \over 3}} \right)} \over 2} + 3\cos 2x + 1 \cr & y = {{3 - {1 \over 2}\cos 2x - {{\sqrt 3 } \over 2}\sin 2x - {1 \over 2}\cos 2x + {{\sqrt 3 } \over 2}\sin 2x - {1 \over 2}\cos 2x - {{\sqrt 3 } \over 2}\sin 2x} \over 2} + 3\cos 2x + 1 \cr & y = {3 \over 2} - {3 \over 4}\cos 2x - {{\sqrt 3 } \over 4}\sin 2x + 3\cos 2x + 1 \cr & y = - {{\sqrt 3 } \over 4}\sin 2x + {9 \over 4}\cos 2x + {5 \over 2} \cr & Ta\,\,co:\,\, - {{\sqrt {21} } \over 2} \le - {{\sqrt 3 } \over 4}\sin 2x + {9 \over 4}\cos 2x \le {{\sqrt {21} } \over 2} \cr & \Rightarrow - {{\sqrt {21} } \over 2} + {5 \over 2} \le y \le {{\sqrt {21} } \over 2} + {5 \over 2} \cr & \Rightarrow \min y = - {{\sqrt {21} } \over 2} + {5 \over 2};\,\,\max y = {{\sqrt {21} } \over 2} + {5 \over 2} \cr} $$