You should not have divided by $\sin x$. That causes you to lose information you need to solve the problem.
\begin{align*}
\sin(2x) & > \sqrt{2}\sin x\\
2\sin x\cos x & > \sqrt{2}\sin x\\
2\sin x\cos x - \sqrt{2}\sin x & > 0\\
\sqrt{2}\sin x(\sqrt{2}\cos x - 1) & > 0
\end{align*}
The inequality is satisfied if $\sin x > 0$ and $\sqrt{2}\cos x - 1 > 0$ or if $\sin x < 0$ and $\sqrt{2}\cos x - 1 < 0$.
Let's focus on solving the problem in the interval $[0, 2\pi)$ for the moment.
The inequality $\sin x > 0$ is satisfied in the interval $[0, 2\pi)$ if $x \in (0, \pi)$. The inequality $$\sqrt{2}\cos x - 1 > 0 \iff \cos x > \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$ is satisfied in the interval $[0, 2\pi)$ if $x \in [0, \frac{\pi}{4}) \cup (\frac{7\pi}{4}, 2\pi)$. Hence, $\sin x > 0$ and $\sqrt{2}\cos x - 1 > 0$ if
$$x \in (0, \pi) \cap \left\{\left[0, \frac{\pi}{4}\right) \cup \left(\frac{7\pi}{4}, 2\pi\right)\right\} = \left(0, \frac{\pi}{4}\right)$$
The inequality $\sin x < 0$ is satisfied in the interval $[0, 2\pi)$ if $x \in (\pi, 2\pi)$. The inequality $$\sqrt{2}\cos x - 1 < 0 \iff \cos x < \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$ is satisfied in the interval $[0, 2\pi)$ if $x \in (\frac{\pi}{4}, \frac{7\pi}{4})$. Hence, $\sin x < 0$ and $\sqrt{2}\cos x - 1 < 0$ if
$$x \in (\pi, 2\pi) \cap \left(\frac{\pi}{4}, \frac{7\pi}{4}\right) = \left(\pi, \frac{7\pi}{4}\right)$$
Thus, $\sin(2x) > \sqrt{2}\sin x$ in the interval $[0, 2\pi)$ if
$$x \in \left(0, \frac{\pi}{4}\right) \cup \left(\pi, \frac{7\pi}{4}\right)$$
Since the sine function has period $2\pi$, the general solution is
$$x \in \bigcup_{k \in \mathbb{Z}} \left\{\left(2k\pi, \frac{\pi}{4} + 2k\pi\right) \cup \left(\pi + 2k\pi, \frac{7\pi}{4} + 2k\pi\right)\right\}$$