Ta có:$\cos 2x+\cos\frac{3x}{4}-2=0$$\Leftrightarrow \left\{\begin{array}{l}\cos2x=1\\\cos\frac{3x}{4}=1\end{array}\right.$$\Leftrightarrow \left\{\begin{array}{l}2x=k2\pi, k\in\mathbb{Z}\\\frac{3x}{4}=l2\pi,l\in\mathbb{Z}\end{array}\right.$$\Leftrightarrow \left[\begin{array}{l}x=k\pi\\x=\frac{k8\pi}{3}\end{array}\right.(k\in\mathbb{Z})$
Ta có:
$\cos 2x\le1,\cos\frac{3x}{4}\le1$ nên: $\cos 2x+\cos\frac{3x}{4}-2=0$$\Leftrightarrow \left\{\begin{array}{l}\cos2x=1\\\cos\frac{3x}{4}=1\end{array}\right.$$\Leftrightarrow \left\{\begin{array}{l}2x=k2\pi, k\in\mathbb{Z}\\\frac{3x}{4}=l2\pi,l\in\mathbb{Z}\end{array}\right.$$\Leftrightarrow \left[\begin{array}{l}x=k\pi\\x=\frac{k8\pi}{3}\end{array}\right.(k\in\mathbb{Z})$