8.183. $\frac{6 \cos ^{3} 2 t+2 \sin ^{3} 2 t}{3 \cos 2 t-\sin 2 t}=\cos 4 t$.

8.183. $\frac{6 \cos ^{3} 2 t+2 \sin ^{3} 2 t}{3 \cos 2 t-\sin 2 t}=\cos 4 t$. 

The equation above is already in English and does not require translation. However, if you meant to translate the problem statement into English, it would be:

8.183. $\frac{6 \cos ^{3} 2 t+2 \sin ^{3} 2 t}{3 \cos 2 t-\sin 2 t}=\cos 4 t$.

This equation is already in a mathematical form that is universal and does not need translation. If you need a verbal description of the equation, it would be:

8.183. The fraction (6 times the cosine of 2t cubed plus 2 times the sine of 2t cubed) divided by (3 times the cosine of 2t minus the sine of 2t) equals the cosine of 4t.
## Solution.

Domain of definition: $3 \cos 2 t-\sin 2 t \neq 0$.

From the condition we have

$$
\begin{aligned}
& \frac{6 \cos ^{3} 2 t+2 \sin ^{3} 2 t}{3 \cos 2 t-\sin 2 t}-\cos ^{2} 2 t+\sin ^{2} 2 t=0 . \\
& 6 \cos ^{3} 2 t+2 \sin ^{3} 2 t-3 \cos ^{3} 2 t+3 \cos 2 t \sin ^{2} 2 t+\sin 2 t \cos ^{2} 2 t-\sin ^{3} 2 t=0 \Leftrightarrow \\
& \Leftrightarrow\left(3 \cos ^{3} 2 t+3 \cos 2 t \sin ^{2} 2 t\right)+\left(\sin ^{3} 2 t+\sin 2 t \cos ^{2} 2 t\right)=0 \Leftrightarrow 3 \cos 2 t\left(\cos ^{2} 2 t+\right. \\
& \left.+\sin ^{2} 2 t\right)+\sin 2 t\left(\sin ^{2} 2 t+\cos ^{2} 2 t\right)=0,3 \cos 2 t+\sin 2 t=0 . \operatorname{tg} 2 t=-3, \\
& 2 t=-\operatorname{arctg} 3+\pi k, t=-\frac{1}{2} \operatorname{arctg} 3+\frac{\pi k}{2}, k \in Z .
\end{aligned}
$$

Answer: $t=-\frac{1}{2} \operatorname{arctg} 3+\frac{\pi k}{2}, k \in Z$.