If `A=[{:(cosx,-sin x),(sinx, cosx):}] and A+A'=I_(2)`, then the general value of x is

Options

A. `n pi, n in I`
B. `(2n+1)(pi)/(2), ne in I`
C. `2n pi +-(pi)/(6)n in I`
D. `2n pi+-(pi)/(3), n in I`

Answer

To solve the problem step by step, we start with the given matrix \( A \) and the condition \( A + A' = I_2 \).

### Step 1: Define the matrix \( A \)
The matrix \( A \) is given as:
\[ A = \begin{pmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{pmatrix} \]

### Step 2: Find the transpose of \( A \)
The transpose of matrix \( A \), denoted as \( A' \), is obtained by interchanging rows and columns:
\[ A' = \begin{pmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{pmatrix} \]

### Step 3: Add \( A \) and \( A' \)
Now, we add \( A \) and \( A' \):
\[ A + A' = \begin{pmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{pmatrix} + \begin{pmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{pmatrix} \]
Calculating the sum:
\[ A + A' = \begin{pmatrix} \cos x + \cos x & -\sin x + \sin x \\ \sin x - \sin x & \cos x + \cos x \end{pmatrix} = \begin{pmatrix} 2\cos x & 0 \\ 0 & 2\cos x \end{pmatrix} \]

### Step 4: Set the sum equal to the identity matrix
According to the problem, we have:
\[ A + A' = I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]
Thus, we set:
\[ \begin{pmatrix} 2\cos x & 0 \\ 0 & 2\cos x \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

### Step 5: Compare corresponding elements
From the above equation, we can compare the corresponding elements:
\[ 2\cos x = 1 \]

### Step 6: Solve for \( \cos x \)
Dividing both sides by 2 gives:
\[ \cos x = \frac{1}{2} \]

### Step 7: Find the general solution for \( x \)
The general solution for \( \cos x = \frac{1}{2} \) is:
\[ x = 2n\pi \pm \frac{\pi}{3}, \quad n \in \mathbb{Z} \]

### Final Answer
Thus, the general value of \( x \) is:
\[ x = 2n\pi + \frac{\pi}{3} \quad \text{or} \quad x = 2n\pi - \frac{\pi}{3}, \quad n \in \mathbb{Z} \]