Given: $A = \begin{bmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{bmatrix}$

The transpose of A, denoted as $A'$, is obtained by interchanging rows and columns:

$A' = \begin{bmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{bmatrix}$

Now, we are given that $A + A' = \sqrt{3}I$, where $I$ is the identity matrix.

So, $\begin{bmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{bmatrix} + \begin{bmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{bmatrix} = \sqrt{3} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Adding the matrices on the left side, we get:

$\begin{bmatrix} 2\cos x & 0 \\ 0 & 2\cos x \end{bmatrix} = \begin{bmatrix} \sqrt{3} & 0 \\ 0 & \sqrt{3} \end{bmatrix}$

Comparing the elements of the matrices, we have:

$2\cos x = \sqrt{3}$

$\cos x = \frac{\sqrt{3}}{2}$

Since $x \in [0, \frac{\pi}{2}]$, we need to find the value of $x$ in this interval for which $\cos x = \frac{\sqrt{3}}{2}$.

We know that $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$.

Therefore, $x = \frac{\pi}{6}$.