$8.446 \tan x + \cot x - \cos 4x = 3$.
Solution.

$\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}=2+(1+\cos 4 x) \Leftrightarrow \frac{2}{\sin 2 x}=2+2 \cos ^{2} 2 x \Leftrightarrow$
$\Leftrightarrow\left\{\begin{array}{l}\sin 2 x \neq 0, \\ \sin 2 x+\cos ^{2} 2 x \cdot \sin 2 x=1\end{array} \Leftrightarrow \sin ^{3} 2 x-2 \sin 2 x+1=0 \Leftrightarrow\right.$

$\Leftrightarrow(\sin 2 x-1)\left(\sin ^{2} 2 x+\sin 2 x-1\right)=0 \Leftrightarrow\left[\begin{array}{l}\sin 2 x=1, \\ \sin 2 x=\frac{-1 \pm \sqrt{5}}{2}\end{array} \Leftrightarrow\right.$

$\Leftrightarrow\left[\begin{array}{l}x=\frac{\pi}{4}+\pi k, \\ x=\frac{1}{2}(-1)^{k} \arcsin \frac{\sqrt{5}-1}{2}+\frac{\pi k}{2} .\end{array}\right.$

Answer: $x_{1}=\frac{\pi}{4}(4 k+1), x_{2}=\frac{1}{2}(-1)^{k} \arcsin \frac{\sqrt{5}-1}{2}+\frac{\pi k}{2}$.