Given $A = \begin{bmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{bmatrix}$. The transpose $A'$ is obtained by interchanging rows and columns: $$A' = \begin{bmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{bmatrix}$$
Adding $A$ and $A'$: $$A + A' = \begin{bmatrix} \cos x + \cos x & -\sin x + \sin x \\ \sin x - \sin x & \cos x + \cos x \end{bmatrix} = \begin{bmatrix} 2\cos x & 0 \\ 0 & 2\cos x \end{bmatrix}$$ Given $A + A' = I$, where $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, we equate the matrices: $$\begin{bmatrix} 2\cos x & 0 \\ 0 & 2\cos x \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
Equating the corresponding elements, we get: $$2\cos x = 1 \implies \cos x = \frac{1}{2}$$ For $x \in [0, \frac{\pi}{2}]$, the value of $x$ that satisfies $\cos x = \frac{1}{2}$ is $x = \frac{\pi}{3}$.
Final Answer: $\frac{\pi}{3}$
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