1. Find all roots of the equation $\frac{1}{\cos ^{3} x}-\frac{1}{\sin ^{3} x}=4 \sqrt{2}$, lying in the interval $\left(-\frac{\pi}{2}, 0\right)$. Write the answer in degrees.
# Solution:

$\sin ^{3} x-\cos ^{3} x=4 \sqrt{2} \sin ^{3} x \cos ^{3} x \Leftrightarrow(\sin x-\cos x)\left(\sin ^{2} x+\sin x \cos x+\cos ^{2} x\right)=4 \sqrt{2} \sin ^{3} x \cos ^{3} x$. Substitution: $\sin x-\cos x=t, \sin x \cos x=\frac{1-t^{2}}{2}$. Then $t\left(3-t^{2}\right)=\sqrt{2}\left(1-t^{2}\right)^{3}$. Substitution: $t=z \sqrt{2}$. The equation will take the form $z\left(3-2 z^{2}\right)-\left(1-2 z^{2}\right)^{3}=0$. There is a root $z=-1$, and the left side can be factored

$$
(z+1)\left(8 z^{5}-8 z^{4}-4 z^{3}+2 z^{2}+4 z-1\right)=0
$$

Since $x \in\left(-\frac{\pi}{2}, 0\right)$, then $t=\sin x-\cos x=\sqrt{2} \sin \left(x-\frac{\pi}{4}\right)\left|4 z^{3}\right|$ and $\left|8 z^{4}\right|>\left|2 z^{2}\right|$. Therefore, $z=-1$ is the only root of equation (1). It is easy to find that $\sin \left(x-\frac{\pi}{4}\right)=-1$, and $x=-\frac{\pi}{4}$.

Answer: -45