Pt\(\Leftrightarrow\dfrac{1}{2}\left(cos2004x+cos2006x\right)=1\)
\(\Leftrightarrow cos2004x+cos2006x=2\)
Có \(\left\{{}\begin{matrix}cos2004x\le1\\cos2006x\le1\end{matrix}\right.\)\(\Rightarrow cos2004x+cos2006x\le2\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}cos2004x=1\\cos2006x=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{k\pi}{1002}\\x=\dfrac{k\pi}{1003}\end{matrix}\right.\)\(\Rightarrow\dfrac{k\pi}{1002}=\dfrac{k\pi}{1003}\Leftrightarrow k=0\Rightarrow x=0\)
Vậy x=0 (Người chơi hệ đêm khuya à)
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