Tìm minmax..y=cos^2 (x+pi/3)+cos^2(x-pi/3)+cos^2(x-2pi/3)+3cos2x +1

Tìm minmax..y=cos^2 (x+pi/3)+cos^2(x-pi/3)+cos^2(x-2pi/3)+3cos2x +1

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

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