1. In the field of real numbers, solve the equation

$$
\sqrt{2}(\sin t+\cos t)=\tan^{3} t+\cot^{3} t .
$$
SOLUTION. From the properties of the tangent and cotangent functions, it follows that $t \neq k \cdot \frac{1}{2} \pi$, where $k$ is any integer. Let us further denote

$$
L=\sqrt{2}(\sin t+\cos t) \quad \text { and } \quad P=\operatorname{tg}^{3} t+\operatorname{cotg}^{3} t
$$

Considering the periodicity of the functions sin, cos, $\operatorname{tg}$, cotg, it suffices to distinguish the following cases.

$\triangleright t \in\left(0 ; \frac{1}{2} \pi\right)$ : For each such $t$, according to the guiding problems 1 and 2, the inequalities hold

$$
\sin t+\cos t \leqq \sqrt{2} \quad \text { and } \quad \operatorname{tg}^{3} t+\operatorname{cotg}^{3} t=\operatorname{tg}^{3} t+\frac{1}{\operatorname{tg}^{3} t} \geqq 2
$$

Equality in each of them occurs precisely when $t=\frac{1}{4} \pi$. We thus obtain the estimate

$$
L \leqq \sqrt{2} \cdot \sqrt{2}=2 \leqq P
$$

The equation $L=P$ is therefore satisfied only in the case $L=P=2$ and the only real number $t$ from the considered interval $\left(0 ; \frac{1}{2} \pi\right)$ that satisfies the given equation is $t=\frac{1}{4} \pi$.

$\triangleright t \in\left(\frac{1}{2} \pi ; \pi\right)$ : For each such $t$, the inequalities hold

$$
\sin t+\cos t>\cos t>-1 \quad \text { and } \quad \operatorname{tg}^{3} t+\operatorname{cotg}^{3} t=-\left((-\operatorname{tg} t)^{3}+\frac{1}{(-\operatorname{tg} t)^{3}}\right) \leqq-2
$$

For any $t$ from the considered interval, the estimates then hold

$$
L>-\sqrt{2}>-2 \geqq P
$$

which means that the given equation has no solutions on this interval.

$>t \in\left(\pi ; \frac{3}{2} \pi\right)$ : For any $t$ from the considered interval, both values $\sin t, \cos t$ are negative (and the values $\operatorname{tg} t, \operatorname{cotg} t$ are positive), so the inequalities hold

$$
\sin t+\cos t<0 \text {. }
$$

Thus,

$$
L<-\sqrt{2} \quad \text { and } \quad \operatorname{tg}^{3} t+\operatorname{cotg}^{3} t \leqq-2
$$

Therefore,

$$
L<-\sqrt{2}>-2 \geqq P
$$

which means that the given equation has no solutions in this case either.

Conclusion. Considering the periodicity of the trigonometric functions involved, the solutions to the given equation are all real numbers $t$ of the form

$$
t=\frac{1}{4} \pi+2 k \pi
$$

where $k$ is any integer.

GUIDING AND SUPPLEMENTARY PROBLEMS: