Step 1: Problem Statement:
The objective is to determine the general value of the angle \( \theta \) by solving a matrix equation involving matrix \( A \), its transpose \( A^T \), and the identity matrix \( I \).
Step 2: Methodology:
1. Compute the transpose of matrix \( A \), denoted as \( A^T \).
2. Calculate the sum \( A^T + A \).
3. Equate the resultant sum to the identity matrix \( I \).
4. Solve the derived trigonometric equation for \( \theta \).
Step 3: Derivation:
Given matrix \( A \) is \( A = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \).
The transpose of \( A \) is obtained by swapping rows and columns:
\[ A^T = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}. \]
The sum \( A^T + A \) is calculated as:
\[ A^T + A = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} + \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \] \[ A^T + A = \begin{bmatrix} 2\cos\theta & 0 \\ 0 & 2\cos\theta \end{bmatrix}. \]
The problem states that \( A^T + A = I \):
\[ \begin{bmatrix} 2\cos\theta & 0 \\ 0 & 2\cos\theta \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. \]
Equating corresponding matrix elements yields:
\[ 2\cos\theta = 1 \] \[ \cos\theta = \frac{1}{2} \]
The principal value for \( \theta \) satisfying \( \cos\theta = \frac{1}{2} \) is \( \frac{\pi}{3} \).
The general solution for the equation \( \cos\theta = \cos\alpha \) is given by \( \theta = 2n\pi \pm \alpha \), where \( n \) is an integer.
Therefore, the general solution for \( \cos\theta = \frac{1}{2} \) is \( \theta = 2n\pi \pm \frac{\pi}{3} \), where \( n \in \mathbb{Z} \).
Considering the provided options, \( \theta = 2n\pi + \frac{\pi}{3} \) represents one component of the general solution and is the most fitting choice.
Step 4: Conclusion:
The general solution for \( \theta \) is \( \theta = 2n\pi + \frac{\pi}{3} \), where \( n \in \mathbb{Z} \).