Bài 1:$4.cos^4x - cos2x - \frac{cos4x}{2} + cos\frac{3x}{4} = \frac{7}{2}$ (1)
Ta có:
$4{\cos ^4}x = {(2{\cos ^2}x)^2} = {(1 + c{\rm{os}}2x)^2} = 1 + 2\cos 2x + \frac{{1 + c{\rm{os}}4x}}{2} = \frac{3}{2} + 2\cos 2x + \frac{{c{\rm{os}}4x}}{2}$
$(1) \Leftrightarrow \frac{3}{2} + \cos 2x + c{\rm{os}}\frac{{3x}}{4} = \frac{7}{2}$
$\Leftrightarrow \cos 2x + c{\rm{os}}\frac{{3x}}{4} = 2$
$ \Leftrightarrow \begin{cases}c{\rm{os}}2x = 1 \\ c{\rm{os}}\frac{{3x}}{4} = 1 \end{cases}$
$ \Leftrightarrow \begin{cases}2x = m2\pi \\ \frac{{3x}}{4} = n2\pi \end{cases} $ ($m,n \in Z$)
$ \Leftrightarrow x = k8\pi (k \in Z)$