The general solution of equation `Sigma_(r=1)^(n) cos(r^(2)theta)sin(rtheta)=(1)/(2)` is

Options

A. `(4K-1)/(n(n+1))(pi)/(2),KinZ`
B. `(2K+1)/(n(n+1))(pi)/(2),KinZ`
C. `(4K+1)/(n(n+1))(pi)/(2),KinZ`
D. None

Explanation

To solve the equation

\[ \sum_{r=1}^{n} \cos(r^2 \theta) \sin(r \theta) = \frac{1}{2}, \]

we will follow a systematic approach.

### Step 1: Use the Product-to-Sum Identity
We start by applying the product-to-sum identities. The identity we will use is:

\[ \cos A \sin B = \frac{1}{2} [\sin(A + B) - \sin(A - B)]. \]

In our case, let \( A = r^2 \theta \) and \( B = r \theta \). Thus, we can rewrite the summation:

\[ \cos(r^2 \theta) \sin(r \theta) = \frac{1}{2} [\sin((r^2 + r)\theta) - \sin((r^2 - r)\theta)]. \]

### Step 2: Substitute into the Summation
Substituting this back into the summation gives:

\[ \sum_{r=1}^{n} \cos(r^2 \theta) \sin(r \theta) = \frac{1}{2} \sum_{r=1}^{n} [\sin((r^2 + r)\theta) - \sin((r^2 - r)\theta)]. \]

### Step 3: Split the Summation
We can split the summation into two parts:

\[ \frac{1}{2} \left( \sum_{r=1}^{n} \sin((r^2 + r)\theta) - \sum_{r=1}^{n} \sin((r^2 - r)\theta) \right) = \frac{1}{2}. \]

### Step 4: Simplify the Expression
Now we need to analyze the two sums. Notice that:

1. The first sum involves \( \sin((r^2 + r)\theta) \).
2. The second sum involves \( \sin((r^2 - r)\theta) \).

### Step 5: Evaluate the Sums
To evaluate these sums, we can use the properties of sine functions and their periodic nature. The key is to find the values of \( \theta \) such that the entire expression equals \( \frac{1}{2} \).

### Step 6: Set Up the Equation
From the equation we have:

\[ \frac{1}{2} \left( \sum_{r=1}^{n} \sin((r^2 + r)\theta) - \sum_{r=1}^{n} \sin((r^2 - r)\theta) \right) = \frac{1}{2}. \]

This simplifies to:

\[ \sum_{r=1}^{n} \sin((r^2 + r)\theta) - \sum_{r=1}^{n} \sin((r^2 - r)\theta) = 1. \]

### Step 7: Solve for \( \theta \)
To solve for \( \theta \), we can set:

\[ \sin(n^2 \theta + n \theta) = 4k + 1 \cdot \frac{\pi}{2}, \]

where \( k \) is an integer. This leads to:

\[ n^2 \theta + n \theta = (4k + 1) \frac{\pi}{2}. \]

### Step 8: Isolate \( \theta \)
Rearranging gives:

\[ \theta = \frac{(4k + 1) \frac{\pi}{2}}{n(n + 1)}. \]

### Final Solution
Thus, the general solution for \( \theta \) is:

\[ \theta = \frac{(4k + 1) \pi}{2n(n + 1)}, \quad k \in \mathbb{Z}. \]