The equatin `2x = (2n +1)pi (1 - cos x) ,` (where n is a positive integer)

Options

A. has infinitely many real roots
B. has exactly one root
C. has exactly `2n +2` real roots
D. has exactly `2n +3` real roots

Answer

To solve the equation \( 2x = (2n + 1)\pi (1 - \cos x) \), where \( n \) is a positive integer, we will analyze the behavior of both sides of the equation and find the number of intersections, which correspond to the roots of the equation.

### Step-by-Step Solution:

1. **Rewrite the Equation**:
We start with the equation:
\[ 2x = (2n + 1)\pi (1 - \cos x) \]
We can rearrange it as:
\[ \frac{2x}{(2n + 1)\pi} = 1 - \cos x \]

2. **Understanding the Functions**:
The left-hand side of the equation, \( \frac{2x}{(2n + 1)\pi} \), is a linear function that increases without bound as \( x \) increases. The right-hand side, \( 1 - \cos x \), oscillates between 0 and 2, achieving its maximum value of 2 at odd multiples of \( \pi \) and its minimum value of 0 at even multiples of \( \pi \).

3. **Finding Intersections**:
- The function \( 1 - \cos x \) has zeros at \( x = 2k\pi \) (where \( k \) is an integer) and reaches its maximum value of 2 at \( x = (2k + 1)\pi \).
- The linear function \( \frac{2x}{(2n + 1)\pi} \) will intersect the oscillating function \( 1 - \cos x \) at various points.

4. **Identifying Roots**:
- The maximum value of \( 1 - \cos x \) is 2, which occurs at odd multiples of \( \pi \).
- The linear function will intersect the oscillating function at points where \( \frac{2x}{(2n + 1)\pi} \) is less than or equal to 2. This condition gives:
\[ x \leq \frac{(2n + 1)\pi}{2} \cdot 2 = (2n + 1)\pi \]
- The intersections will occur at \( x = 0, 2\pi, 4\pi, \ldots, (2n + 1)\pi \).

5. **Counting the Roots**:
- The number of intersections (roots) will be determined by the number of times the oscillating function crosses the linear function from 0 to \( (2n + 1)\pi \).
- The intersections occur at \( 0, 2\pi, 4\pi, \ldots, (2n + 1)\pi \), which gives us \( n + 1 \) points.
- Additionally, for each odd multiple of \( \pi \) (i.e., \( \pi, 3\pi, 5\pi, \ldots, (2n + 1)\pi \)), there will be additional intersections.

6. **Final Count**:
- The total number of intersections is \( 2n + 2 \) (including the intersections at even multiples of \( \pi \) and odd multiples of \( \pi \)).
- Therefore, the equation \( 2x = (2n + 1)\pi (1 - \cos x) \) has exactly \( 2n + 2 \) real roots.

### Conclusion:
Thus, the answer is that the equation has exactly \( 2n + 2 \) real roots.