8.303. $\left(\cos ^{-6} z-\operatorname{tg}^{6} z-\frac{7}{3}\right) \cdot(\sin z+\cos z+2)=0$. 

Translate the above text into English, keeping the original text's line breaks and format:

8.303. $\left(\cos ^{-6} z-\tan^{6} z-\frac{7}{3}\right) \cdot(\sin z+\cos z+2)=0$.
## Solution.

Domain of definition: $\cos z \neq 0$.

From the condition we have

$$
\begin{aligned}
& {\left[\frac{1}{\cos ^{6} z}-\frac{\sin ^{6} z}{\cos ^{6} z}-\frac{7}{3}=0, \Leftrightarrow 3-3 \sin ^{6} z-7 \cos ^{6} z=0 \Leftrightarrow\right.} \\
& \Leftrightarrow 3-3\left(\sin ^{2} z\right)^{3}-7\left(\cos ^{2} z\right)^{3}=0 \Leftrightarrow \\
& \Leftrightarrow 3-3\left(\frac{1}{2}(1-\cos 2 z)\right)^{3}-7\left(\frac{1}{2}(1+\cos 2 z)\right)^{3}=0 \Leftrightarrow \\
& \Leftrightarrow 2 \cos ^{3} 2 z+15 \cos ^{2} 2 z+6 \cos 2 z-7=0 \Leftrightarrow \\
& \Leftrightarrow 2 \cos ^{3} 2 z+2 \cos ^{2} 2 z+6 \cos ^{2} 2 z+6 \cos 2 z+7 \cos ^{2} 2 z-7=0 \Leftrightarrow \\
& \Leftrightarrow 2 \cos ^{2} 2 z(\cos 2 z+1)+6 \cos 2 z(\cos 2 z+1)+7(\cos 2 z+1)(\cos 2 z-1)=0 \\
& (\cos 2 z+1)\left(2 \cos ^{2} 2 z+13 \cos 2 z-7\right)=0
\end{aligned}
$$

From this,

1) $\cos 2 z+1=0, \cos 2 z=-1, 2 z=\pi+2 \pi k, z_{1}=\frac{\pi}{2}+\pi k, k \in Z$;
2) $2 \cos ^{2} 2 z+13 \cos 2 z-7=0$.

Solving the last equation as a quadratic in $\cos 2 z$, we get $\cos 2 z=-7, \varnothing ; \cos 2 z=\frac{1}{2}, 2 z= \pm \frac{\pi}{3}+2 \pi n, z_{2}= \pm \frac{\pi}{6}+\pi n=\frac{\pi}{6}(6 n \pm 1)$, $n \in Z ; z_{1}$ does not satisfy the domain of definition.

Answer: $z=\frac{\pi}{6}(6 n \pm 1), n \in Z$.