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To determine the range of $\cos x$ over the interval $\left(-\frac{\pi}{3}, \frac{\pi}{3}\right)$:

• We evaluate $\cos x$ at the endpoints: $\cos\left(-\frac{\pi}{3}\right) = \frac{1}{2}$ and $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$.
• We also consider any critical points (where the derivative is zero or undefined) within the interval. For $\cos x$, its derivative is $-\sin x$, which is 0 at $x=0$.
• At $x=0$, $\cos(0) = 1$.

Since the interval is open, $x$ cannot be exactly $-\frac{\pi}{3}$ or $\frac{\pi}{3}$, so $\cos x$ will be strictly greater than $\frac{1}{2}$. However, $x=0$ is included in the interval, so $\cos x$ can be equal to 1.
Therefore, the range of $\cos x$ for $x \in \left(-\frac{\pi}{3}, \frac{\pi}{3}\right)$ is $\frac{1}{2} < \cos x \le 1$. Considering $x=0$ was crucial to establish the upper bound of 1 for $\cos x$.