192*. The sequence $\left(x_{n}\right)$ is defined by the conditions:

$$
x_{1}=\frac{1}{2}, \quad x_{n+1}=\sqrt{\frac{1-\sqrt{1-x_{n}^{2}}}{2}}
$$

Find a formula for $x_{n}$.
$\triangle$ From the condition, it is clear that $0 \leq x_{n} \leq 1$. Let $x_{n}=\sin \alpha$, where $\alpha \in\left[0 ; \frac{\pi}{2}\right]$. We get:

$$
x_{n+1}=\sqrt{\frac{1-\sqrt{1-\sin ^{2} \alpha}}{2}}=\sqrt{\frac{1-\cos \alpha}{2}}=\sin \frac{\alpha}{2}
$$

The argument of $x_{n+1}$ under the sine sign turned out to be twice as small as that of $x_{n}$. Then, since

$$
x_{1}=\sin \frac{\pi}{6}=\sin \frac{\pi}{3 \cdot 2}
$$

the argument of $x_{2}$ under the sine sign is $\frac{\pi}{3 \cdot 2^{2}}, \mathrm{y} x_{3}-\frac{\pi}{3 \cdot 2^{3}}$, generally, the argument of $x_{n}-\frac{\pi}{3 \cdot 2^{n}}$.

Answer: $x_{n}=\sin \frac{\pi}{3 \cdot 2^{n}}$.