1. The problem is to find the values of $x$ for which $\cos x < 0$. 2. Recall that the cosine function $\cos x$ is negative in the intervals where the angle $x$ lies in the second and third quadrants of the unit circle. 3. The cosine function has period $2\pi$, so the intervals repeat every $2\pi$. 4. Specifically, $\cos x < 0$ when: $$\pi/2 < x < 3\pi/2 + 2k\pi, \quad k \in \mathbb{Z}$$ 5. This means the solution set is all $x$ such that: $$x \in \left(\pi/2 + 2k\pi, 3\pi/2 + 2k\pi\right), \quad k \in \mathbb{Z}$$ 6. In plain language, cosine is negative between $90^\circ$ and $270^\circ$ (or $\pi/2$ and $3\pi/2$ radians) plus any full rotations of $2\pi$ radians. Final answer: $$x \in \bigcup_{k=-\infty}^{\infty} \left(\pi/2 + 2k\pi, 3\pi/2 + 2k\pi\right)$$