$\frac{{c{\rm{os}}\frac{{4x}}{3} - cos^2x}}{{\sqrt {1 - {{\tan }^2}x} }} = 0\Leftrightarrow \begin{cases}cos\frac{4x}{3}=cos^2x \\ 1-\frac{sin^2x}{cos^2x}>0 \end{cases} \Leftrightarrow \begin{cases}cos\frac{4x}{3}=\frac{1+cos2x}{2} \\ \frac{cos2x}{cos^2x} >0 \end{cases} $
$\Leftrightarrow \begin{cases}2cos\frac{4x}{3}=1+cos2x \\ cos2x>0 \\cos^2x\neq 0\end{cases} \Leftrightarrow \begin{cases}2(2cos^2\frac{2x}{3} -1)=1+4cos^3\frac{2x}{3}-3cos\frac{2x}{3} \\ cos2x>0 \\\frac{1+cos2x}{2}\neq 0\Leftrightarrow cos2x\neq -1 \end{cases} $
Đặt $cos\frac{2x}{3}=t $ ta được:
$4t^3-4t^2-3t+3=0 \Leftrightarrow$$\left[\begin{array}{I}
t=1\\t=\pm\frac{\sqrt{3}}{2}\end{array}\right.\Leftrightarrow\left[\begin{array}{I}\cos\frac{2x}{3}=1\\\cos\frac{2x}{3}=\pm\frac{\sqrt{3}}{2}\end{array}\right.$
$\Leftrightarrow\left[\begin{array}{I}
x=3k\pi\\x=\pm\frac{\pi}{2}+k3\pi\\x=\pm\frac{5\pi}{4}+k3\pi\end{array}\right.$Kết hợp ĐK
$\cos 2x>0\Rightarrow x=3k\pi, k\in Z$