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Ta có:
$\left\{\begin{array}{l}\cos\dfrac{3x}{4}\le 1\\\cos 2x\le1\end{array}\right.\Rightarrow \cos\dfrac{3x}{4}+\cos2x\le 2$
Dấu bằng xảy ra khi:
$\left\{\begin{array}{l}\cos\dfrac{3x}{4}=1\\\cos 2x=1\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}\dfrac{3x}{4}=k\pi&,k\in\mathbb{Z}\\2x=l\pi&,l\in\mathbb{Z}\end{array}\right.$
$\Leftrightarrow x=k4\pi,k\in\mathbb{Z}$
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