$5.38 \frac{1}{\cos ^{4} z}=\frac{160}{9}-\frac{2(\operatorname{ctg} 2 z \operatorname{ctg} z+1)}{\sin ^{2} z}$.

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$5.38 \frac{1}{\cos ^{4} z}=\frac{160}{9}-\frac{2(\cot 2 z \cot z+1)}{\sin ^{2} z}$.
5.38 Here $\cos z \neq 0, \sin z \neq 0$. Notice that

$\operatorname{ctg} 2 z \operatorname{ctg} z+1=\frac{\cos 2 z \cos z+\sin 2 z \sin z}{\sin 2 z \sin z}=$

$=\frac{\cos z}{2 \sin z \cos z \sin z}=\frac{1}{2 \sin ^{2} z}$.

Then we obtain the equation

$\frac{1}{\sin ^{4} z}+\frac{1}{\cos ^{4} z}=\frac{160}{9} ; \sin ^{4} z+\cos ^{4} z=\frac{160}{9} \sin ^{4} z \cos ^{4} z$.

Since $\sin ^{4} z+\cos ^{4} z+2 \sin ^{2} z \cos ^{2} z=\left(\sin ^{2} z+\cos ^{2} z\right)^{2}=1$, the equation becomes

$$
1-2 \sin ^{2} z \cos ^{2} z=\frac{160}{9} \sin ^{4} z \cos ^{4} z
$$

$20 \sin ^{4} 2 z+9 \sin ^{2} 2 z-18=0 ; 20 u^{2}+9 u-18=0$, where $u=\sin ^{2} 2 z$.

From this, we find $u=\frac{3}{4}$ (taking only the positive root). Thus, $\sin 2 z= \pm \frac{\sqrt{3}}{2}$, i.e., $z=\frac{\pi}{6}(3 k \pm 1)$.

Answer: $z=\frac{\pi}{6}(3 k \pm 1), k \in Z$.