There is one K numbered of Z that verify -π<(π/3)+2kπ<π

We are tasked with finding the values of \( k \) such that the inequality

\[
-\pi < \frac{\pi}{3} + 2k\pi < \pi
\]

is satisfied.

### Step-by-Step Solution:

1. **Break the inequality into two parts**:

   The inequality can be split into two parts:
   
   \[
   -\pi < \frac{\pi}{3} + 2k\pi
   \]
   and
   \[
   \frac{\pi}{3} + 2k\pi < \pi
   \]

2. **Solve the first part**: 

   \[
   -\pi < \frac{\pi}{3} + 2k\pi
   \]

   Subtract \(\frac{\pi}{3}\) from both sides:

   \[
   -\pi - \frac{\pi}{3} < 2k\pi
   \]

   Combine the terms on the left-hand side:

   \[
   -\frac{3\pi}{3} - \frac{\pi}{3} = -\frac{4\pi}{3}
   \]

   Thus, we have:

   \[
   -\frac{4\pi}{3} < 2k\pi
   \]

   Divide both sides by \(2\pi\):

   \[
   -\frac{4}{6} < k
   \]

   Simplifying gives:

   \[
   -\frac{2}{3} < k
   \]

   So \( k > -\frac{2}{3} \).

3. **Solve the second part**:

   \[
   \frac{\pi}{3} + 2k\pi < \pi
   \]

   Subtract \(\frac{\pi}{3}\) from both sides:

   \[
   2k\pi < \pi - \frac{\pi}{3}
   \]

   Combine the terms on the right-hand side:

   \[
   2k\pi < \frac{3\pi}{3} - \frac{\pi}{3} = \frac{2\pi}{3}
   \]

   Divide both sides by \(2\pi\):

   \[
   k < \frac{2}{6}
   \]

   Simplifying gives:

   \[
   k < \frac{1}{3}
   \]

4. **Combine the two results**:

   We now have the two conditions:
   
   \[
   -\frac{2}{3} < k < \frac{1}{3}
   \]

Since \( k \) is an integer, the only integer value that satisfies this inequality is \( k = 0 \).

Thus, the value of \( k \) that satisfies the inequality is:

\[
k = 0
\]

### Would you like more details or have any questions?

Here are five related questions you might find interesting:

1. How would the inequality change if the bounds were different, for example \( -2\pi < \frac{\pi}{3} + 2k\pi < 2\pi \)?
2. What happens to the value of \( k \) if we modify the factor of \( 2k\pi \) to \( 3k\pi \)?
3. How does adding a constant to both sides of an inequality affect the solution set?
4. Can we solve a similar inequality for a trigonometric function like sine or cosine?
5. How would this inequality change if \( k \) were restricted to real numbers instead of integers?

**Tip:** When solving inequalities involving periodic functions like \( \pi \), always consider the periodicity and symmetries of the functions involved.

Learn how to solve the inequality -π < (π/3) + 2kπ < π to find the integer value of k. This problem explores the properties of inequalities and periodic trigonometric functions. Suitable for students in Grades 10-12.

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