Step-by-Step Solution

1. First, calculate \( [F(x)]^2 \):

   \( [F(x)]^2 = F(x) \cdot F(x) = \begin{bmatrix}\cos~x&-\sin~x&0\\ \sin~x&\cos~x&0\\ 0&0&1\end{bmatrix} \cdot \begin{bmatrix}\cos~x&-\sin~x&0\\ \sin~x&\cos~x&0\\ 0&0&1\end{bmatrix} \)

2. Perform the matrix multiplication:

   \( [F(x)]^2 = \begin{bmatrix}\cos^2 x - \sin^2 x&-\cos x \sin x - \sin x \cos x&0\\ \sin x \cos x + \cos x \sin x&-\sin^2 x + \cos^2 x&0\\ 0&0&1\end{bmatrix} \)

3. Simplify using trigonometric identities:

   \( [F(x)]^2 = \begin{bmatrix}\cos 2x&-\sin 2x&0\\ \sin 2x&\cos 2x&0\\ 0&0&1\end{bmatrix} \)

4. We are given that \( [F(x)]^2 = F(kx) \). Therefore:

   \( F(kx) = \begin{bmatrix}\cos kx&-\sin kx&0\\ \sin kx&\cos kx&0\\ 0&0&1\end{bmatrix} \)

5. Comparing \( [F(x)]^2 \) with \( F(kx) \), we have:

   \( \begin{bmatrix}\cos 2x&-\sin 2x&0\\ \sin 2x&\cos 2x&0\\ 0&0&1\end{bmatrix} = \begin{bmatrix}\cos kx&-\sin kx&0\\ \sin kx&\cos kx&0\\ 0&0&1\end{bmatrix} \)

6. By comparing the elements of the matrices, we can deduce that:

   \( \cos 2x = \cos kx \) and \( \sin 2x = \sin kx \)

7. Therefore, \( k = 2 \).