Solve `:(tanx-(1)/(sqrt3))(cosx-(1)/(2))le0.`

Answer

To solve the inequality \((\tan x - \frac{1}{\sqrt{3}})(\cos x - \frac{1}{2}) \leq 0\), we will follow these steps:

### Step 1: Solve \(\tan x - \frac{1}{\sqrt{3}} = 0\)

Set \(\tan x = \frac{1}{\sqrt{3}}\). The angle whose tangent is \(\frac{1}{\sqrt{3}}\) is \(\frac{\pi}{6}\). Thus, we have:
\[ x = \tan^{-1} \left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \]

### Step 2: Solve \(\cos x - \frac{1}{2} = 0\)

Set \(\cos x = \frac{1}{2}\). The angle whose cosine is \(\frac{1}{2}\) is \(\frac{\pi}{3}\). Thus, we have:
\[ x = \cos^{-1} \left(\frac{1}{2}\right) = \frac{\pi}{3} \]

### Step 3: Identify critical points

The critical points from the above steps are:
- \(x = \frac{\pi}{6}\)
- \(x = \frac{\pi}{3}\)

### Step 4: Test intervals

We will test the intervals determined by the critical points \(\frac{\pi}{6}\) and \(\frac{\pi}{3}\):
1. Interval \((-\infty, \frac{\pi}{6})\)
2. Interval \((\frac{\pi}{6}, \frac{\pi}{3})\)
3. Interval \((\frac{\pi}{3}, \infty)\)

Choose test points from each interval:
- For \(x = 0\) (in \((-\infty, \frac{\pi}{6})\)):
\[ \tan(0) - \frac{1}{\sqrt{3}} < 0 \quad \text{and} \quad \cos(0) - \frac{1}{2} > 0 \quad \Rightarrow \quad (-)(+) > 0 \]
- For \(x = \frac{\pi}{4}\) (in \((\frac{\pi}{6}, \frac{\pi}{3})\)):
\[ \tan\left(\frac{\pi}{4}\right) - \frac{1}{\sqrt{3}} > 0 \quad \text{and} \quad \cos\left(\frac{\pi}{4}\right) - \frac{1}{2} > 0 \quad \Rightarrow \quad (+)(+) > 0 \]
- For \(x = \frac{\pi}{2}\) (in \((\frac{\pi}{3}, \infty)\)):
\[ \tan\left(\frac{\pi}{2}\right) \text{ is undefined} \quad \text{and} \quad \cos\left(\frac{\pi}{2}\right) - \frac{1}{2} < 0 \quad \Rightarrow \quad \text{undefined} \text{ and } (-) \text{ does not contribute} \]

### Step 5: Determine the sign of the product

From the tests:
- In \((-\infty, \frac{\pi}{6})\): Positive
- In \((\frac{\pi}{6}, \frac{\pi}{3})\): Negative
- In \((\frac{\pi}{3}, \infty)\): Undefined or negative

### Step 6: Include critical points

Since we need \((\tan x - \frac{1}{\sqrt{3}})(\cos x - \frac{1}{2}) \leq 0\), we include the points where the expression equals zero:
- At \(x = \frac{\pi}{6}\) and \(x = \frac{\pi}{3}\), the expression equals zero.

### Final Solution

Thus, the solution to the inequality is:
\[ x \in \left[\frac{\pi}{6}, \frac{\pi}{3}\right] \]

### General Solution

The general solution can be expressed as:
\[ x = \frac{\pi}{6} + n\pi \quad \text{to} \quad x = \frac{\pi}{3} + n\pi, \quad n \in \mathbb{Z} \]