$8.438 \cos \left(x-\frac{\pi}{4}\right) \cdot\left(1-4 \cos ^{2} 2 x\right)-2 \cos 4 x=3$.
## Solution.

$$
\begin{aligned}
& \cos \left(x-\frac{\pi}{4}\right) \cdot\left(1-4 \cos ^{2} 2 x\right)=1+2(\cos 4 x+1) \Leftrightarrow \\
& \Leftrightarrow \cos \left(x-\frac{\pi}{4}\right) \cdot\left(1-4 \cos ^{2} 2 x\right)=1+4 \cos ^{2} 2 x \Leftrightarrow \cos \left(x-\frac{\pi}{4}\right)=\frac{1+4 \cos ^{2} 2 x}{1-4 \cos ^{2} 2 x}
\end{aligned}
$$

Note that the right-hand side of this equation should not exceed 1 in absolute value, i.e., $\left|\frac{1+4 \cos ^{2} 2 x}{1-4 \cos ^{2} 2 x}\right| \leq 1 \Leftrightarrow 1+4 \cos ^{2} 2 x \leq\left|1-4 \cos ^{2} 2 x\right| \Leftrightarrow$

$\Leftrightarrow\left[\begin{array}{l}1-4 \cos ^{2} 2 x \geq 1+4 \cos ^{2} 2 x, \\ 1-4 \cos ^{2} 2 x \leq-\left(1+4 \cos ^{2} 2 x\right)\end{array} \Leftrightarrow\left[\begin{array}{l}\cos ^{2} 2 x=0, \\ \varnothing\end{array} \Leftrightarrow \cos 2 x=0 \Leftrightarrow x=\frac{\pi}{4}+\frac{\pi n}{2}\right.\right.$.

Substituting these values into the original equation, we have:

$\cos \frac{\pi n}{2}=1 \Rightarrow n=4 k \Rightarrow x=\frac{\pi}{4}+2 \pi k$.

Answer: $x=\frac{\pi}{4}(8 k+1)$.