1. Given $x \in[0,2 \pi]$, and $\sin x=\sin \left[\arcsin \frac{2}{3}-\arcsin \left(-\frac{1}{3}\right)\right], x=$
1. $\arcsin \frac{4 \sqrt{2}+\sqrt{5}}{9}$ or $\pi-\arcsin \frac{4 \sqrt{2}+\sqrt{5}}{9}$.
$$
\sin x=\sin \left(\arcsin \frac{2}{3}+\arcsin \frac{1}{3}\right)=\frac{2}{3} \times \sqrt{1-\left(\frac{1}{3}\right)^{2}}+\sqrt{1-\left(\frac{2}{3}\right)^{2}} \times \frac{1}{3}=\frac{4 \sqrt{2}+\sqrt{5}}{9}
$$

So $x=\arcsin \frac{4 \sqrt{2}+\sqrt{5}}{9}$ or $x=\pi-\arcsin \frac{4 \sqrt{2}+\sqrt{5}}{9}$.