1. The problem is to find the values of $x$ for which $\cos x > -\frac{\sqrt{2}}{2}$.\n\n2. Recall that $\cos x$ ranges between $-1$ and $1$. The value $-\frac{\sqrt{2}}{2}$ is approximately $-0.707$. We want to find where the cosine function is greater than this value.\n\n3. The cosine function equals $-\frac{\sqrt{2}}{2}$ at angles $x = \frac{3\pi}{4} + 2k\pi$ and $x = \frac{5\pi}{4} + 2k\pi$ for any integer $k$.\n\n4. Since cosine decreases from 1 at $0$ to $-1$ at $\pi$, and then increases back to 1 at $2\pi$, the inequality $\cos x > -\frac{\sqrt{2}}{2}$ holds outside the interval $[\frac{3\pi}{4}, \frac{5\pi}{4}]$ within one period.\n\n5. Therefore, the solution set is all $x$ such that $x \in ( -\infty, \frac{3\pi}{4} + 2k\pi ) \cup ( \frac{5\pi}{4} + 2k\pi, \infty )$ for all integers $k$.\n\n6. In simpler terms, $\cos x$ is greater than $-\frac{\sqrt{2}}{2}$ everywhere except between $\frac{3\pi}{4}$ and $\frac{5\pi}{4}$ plus full rotations of $2\pi$.