`f(x) = {{:((1-coskx)/(x sinx), if x ne 0),(1/2, if x = 0):}` at `x = 0 `

Explanation

To solve the problem, we need to find the value of \( k \) such that the function

\[ f(x) = \begin{cases} \frac{1 - \cos(kx)}{x \sin x} & \text{if } x \neq 0 \\ \frac{1}{2} & \text{if } x = 0 \end{cases} \]

is continuous at \( x = 0 \).

### Step-by-Step Solution:

1. **Understanding Continuity**:
For the function to be continuous at \( x = 0 \), we need to satisfy the condition:
\[ \lim_{x \to 0} f(x) = f(0) \]
Given that \( f(0) = \frac{1}{2} \), we need to find \( \lim_{x \to 0} f(x) \).

2. **Finding the Limit**:
We need to calculate:
\[ \lim_{x \to 0} \frac{1 - \cos(kx)}{x \sin x} \]

3. **Using the Cosine Identity**:
We know that:
\[ 1 - \cos(kx) = 2 \sin^2\left(\frac{kx}{2}\right) \]
Thus, we can rewrite the limit:
\[ \lim_{x \to 0} \frac{2 \sin^2\left(\frac{kx}{2}\right)}{x \sin x} \]

4. **Separating the Limits**:
We can separate the limit as follows:
\[ \lim_{x \to 0} \frac{2 \sin^2\left(\frac{kx}{2}\right)}{x^2} \cdot \frac{x^2}{x \sin x} \]

5. **Applying the Limit**:
We know:
\[ \lim_{x \to 0} \frac{\sin(kx/2)}{kx/2} = 1 \]
Therefore:
\[ \lim_{x \to 0} \frac{\sin^2\left(\frac{kx}{2}\right)}{(kx/2)^2} = 1 \]
Thus, we have:
\[ \lim_{x \to 0} \frac{2 \sin^2\left(\frac{kx}{2}\right)}{x^2} = 2 \cdot \frac{k^2}{4} = \frac{k^2}{2} \]

6. **Finding the Limit of the Denominator**:
We also know:
\[ \lim_{x \to 0} \frac{x}{\sin x} = 1 \]
Therefore:
\[ \lim_{x \to 0} \frac{x^2}{x \sin x} = \lim_{x \to 0} \frac{x}{\sin x} = 1 \]

7. **Combining the Results**:
Now, we can combine the limits:
\[ \lim_{x \to 0} f(x) = \frac{k^2}{2} \cdot 1 = \frac{k^2}{2} \]

8. **Setting the Limits Equal**:
Since we need \( \lim_{x \to 0} f(x) = f(0) \):
\[ \frac{k^2}{2} = \frac{1}{2} \]

9. **Solving for \( k \)**:
Multiply both sides by 2:
\[ k^2 = 1 \]
Taking the square root:
\[ k = \pm 1 \]

### Final Answer:
The values of \( k \) that make the function continuous at \( x = 0 \) are:
\[ k = 1 \quad \text{or} \quad k = -1 \]