1. The problem is to solve the inequality $\sin x < \frac{\sqrt{2}}{2}$.\n\n2. Recall that $\sin x$ is the sine function, which oscillates between $-1$ and $1$. The value $\frac{\sqrt{2}}{2}$ is approximately $0.707$.\n\n3. The sine function equals $\frac{\sqrt{2}}{2}$ at angles $x = \frac{\pi}{4} + 2k\pi$ and $x = \frac{3\pi}{4} + 2k\pi$ for any integer $k$.\n\n4. Since $\sin x$ is less than $\frac{\sqrt{2}}{2}$, the solution includes all $x$ except where $\sin x$ is greater than or equal to $\frac{\sqrt{2}}{2}$.\n\n5. On the interval $[0, 2\pi)$, $\sin x \geq \frac{\sqrt{2}}{2}$$ for $x \in [\frac{\pi}{4}, \frac{3\pi}{4}]$.\n\n6. Therefore, the solution set is all real numbers $x$ such that $x \notin [\frac{\pi}{4} + 2k\pi, \frac{3\pi}{4} + 2k\pi]$ for any integer $k$.\n\n7. In interval notation, the solution is $$\bigcup_{k \in \mathbb{Z}} \left(-\infty, \frac{\pi}{4} + 2k\pi \right) \cup \left(\frac{3\pi}{4} + 2k\pi, \infty \right).$$\n\nThis means $\sin x$ is less than $\frac{\sqrt{2}}{2}$ everywhere except between $\frac{\pi}{4}$ and $\frac{3\pi}{4}$ plus full periods of $2\pi$.