Solve the inequality $\cos(\frac{1}{2}\theta - \frac{\pi}{3}) \le \frac{1}{\sqrt{2}}$.

1. Problem Description

Solve the inequality $\cos(\frac{1}{2}\theta - \frac{\pi}{3}) \le \frac{1}{\sqrt{2}}$.

2. Solution Steps

First, we know that $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.

The inequality $\cos(x) \le \frac{1}{\sqrt{2}}$ implies that $x$ is in the interval $[\frac{\pi}{4}, 2\pi - \frac{\pi}{4}]$ or more generally $[ \frac{\pi}{4} + 2k\pi, \frac{7\pi}{4} + 2k\pi]$ for any integer $k$. Thus we have

$\frac{\pi}{4} + 2k\pi \le \frac{1}{2}\theta - \frac{\pi}{3} \le \frac{7\pi}{4} + 2k\pi$.

Add $\frac{\pi}{3}$ to all parts of the inequality:

$\frac{\pi}{4} + \frac{\pi}{3} + 2k\pi \le \frac{1}{2}\theta \le \frac{7\pi}{4} + \frac{\pi}{3} + 2k\pi$

$\frac{3\pi + 4\pi}{12} + 2k\pi \le \frac{1}{2}\theta \le \frac{21\pi + 4\pi}{12} + 2k\pi$

$\frac{7\pi}{12} + 2k\pi \le \frac{1}{2}\theta \le \frac{25\pi}{12} + 2k\pi$

Multiply all parts of the inequality by 2:

$\frac{7\pi}{6} + 4k\pi \le \theta \le \frac{25\pi}{6} + 4k\pi$.

For $k=0$, we have $\frac{7\pi}{6} \le \theta \le \frac{25\pi}{6}$.

Since $\theta$ is typically defined between $0$ and $2\pi$, we need to find the angles between $0$ and $2\pi$.

We want to express the solution within the interval $[0, 2\pi)$.

The inequality is $\frac{7\pi}{6} + 4k\pi \le \theta \le \frac{25\pi}{6} + 4k\pi$.

If $k=0$, $\frac{7\pi}{6} \le \theta \le \frac{25\pi}{6}$.

However, we need $\theta \in [0, 2\pi)$.

So, $\frac{7\pi}{6} \le \theta \le \frac{25\pi}{6}$

Also note that $\frac{25\pi}{6} = 4\pi + \frac{\pi}{6}$, so $\frac{25\pi}{6}$ is larger than $2\pi$, so we need to consider values when $k = -1$.

If $k=-1$, $\frac{7\pi}{6} - 4\pi \le \theta \le \frac{25\pi}{6} - 4\pi$

$\frac{7\pi - 24\pi}{6} \le \theta \le \frac{25\pi - 24\pi}{6}$

$\frac{-17\pi}{6} \le \theta \le \frac{\pi}{6}$.

Since we want $\theta \in [0, 2\pi)$, $\theta$ must be in the interval $[\frac{7\pi}{6}, 2\pi) \cup [0, \frac{\pi}{6}]$ so consider $k=0$, $\frac{7\pi}{6} \le \theta \le \frac{25\pi}{6}$.

We can consider $\theta \in [0,2\pi]$.

$\frac{7\pi}{6} \le \theta \le 2\pi$, $\frac{7\pi}{6} \le \theta \le \frac{12\pi}{6}$, so $\theta \in [\frac{7\pi}{6}, \frac{12\pi}{6}]$

and $0 \le \theta \le \frac{\pi}{6}$.

Thus $\theta \in [0, \frac{\pi}{6}] \cup [\frac{7\pi}{6}, 2\pi]$.

3. Final Answer

$[0, \frac{\pi}{6}] \cup [\frac{7\pi}{6}, 2\pi]$