Find the solution set of the inequality `cosx>= (-1)/2.`

Answer

To solve the inequality \( \cos x \geq -\frac{1}{2} \), we can follow these steps:

### Step 1: Identify the critical angles
The cosine function equals \(-\frac{1}{2}\) at specific angles. From the unit circle, we know:
- \( \cos \left( \frac{2\pi}{3} \right) = -\frac{1}{2} \)
- \( \cos \left( \frac{4\pi}{3} \right) = -\frac{1}{2} \)

### Step 2: Determine the intervals
Next, we need to find the intervals where \( \cos x \) is greater than or equal to \(-\frac{1}{2}\). The cosine function is greater than or equal to \(-\frac{1}{2}\) in the following intervals:
- From \( 0 \) to \( \frac{2\pi}{3} \) (first quadrant and part of the second quadrant)
- From \( \frac{4\pi}{3} \) to \( 2\pi \) (part of the third quadrant and fourth quadrant)

### Step 3: Write the solution set
Since the cosine function is periodic with a period of \( 2\pi \), we can express the solution set as:
- \( x \in [0, \frac{2\pi}{3}] \cup [\frac{4\pi}{3}, 2\pi] + 2k\pi \) where \( k \) is any integer.

### Final Solution
Thus, the solution set for the inequality \( \cos x \geq -\frac{1}{2} \) is:
\[ x \in [0, \frac{2\pi}{3}] \cup [\frac{4\pi}{3}, 2\pi] + 2k\pi, \quad k \in \mathbb{Z} \]

---