2. Solve the equation $\sqrt{\frac{2 t}{1+t^{2}}}+\sqrt[3]{\frac{1-t^{2}}{1+t^{2}}}=1$.
Solution. Let's make the substitution

$$
t=\operatorname{tg} \frac{\alpha}{2}, \alpha \in(-\pi, \pi)
$$

Then

$$
\frac{2 t}{1+t^{2}}=\sin \alpha, \frac{1-t^{2}}{1+t^{2}}=\cos \alpha
$$

and the original equation will take the form

$$
\sqrt{\sin \alpha}+\sqrt[3]{\cos \alpha}=1
$$

If $\cos \alpha<0$, then the left side of (2) is strictly less than 1, and (2) has no roots. In the case when $\cos \alpha \geq 0$ and $\sin \alpha \geq 0$, we have the obvious inequality

$$
\sqrt{\sin \alpha}+\sqrt[3]{\cos \alpha} \geq \cos ^{2} \alpha+\sin ^{2} \alpha=1
$$

Equality is achieved only when either $\cos \alpha=1$, or $\sin \alpha=1$. Thus, either $\alpha=0$, or $\alpha=\pi / 2$. Substituting the found values of $\alpha$ into (1), we will find the desired $t$.

Answer: $t \in\{0,1\}$.