When working with determinants, especially for \( 3 \times 3 \) matrices, it’s essential to break the determinant into smaller \( 2 \times 2 \) minors for each element in the first row. After expanding, simplify the expressions and evaluate each minor. Additionally, be mindful of trigonometric identities like \( \cos^2 x + \sin^2 x = 1 \), which can help simplify the result. Also, remember to check the bounds for maximizing or minimizing the function (in this case, based on \( \sin^2 x \)).
The determinant is \( \Delta \).
\[\Delta = \begin{vmatrix} 1 & \cos x & 1 \\ -\cos x & 1 & \cos x \\ -1 & -\cos x & 1 \end{vmatrix}.\]
Expanding the determinant yields:
\[ \Delta = 1 \cdot \begin{vmatrix} 1 & \cos x \\ -\cos x & 1 \end{vmatrix} - \cos x \cdot \begin{vmatrix} -\cos x & \cos x \\ -1 & 1 \end{vmatrix} +1 \cdot \begin{vmatrix} -\cos x & 1 \\ -1 & -\cos x \end{vmatrix}.\]
The minors are calculated as:
\[ \begin{vmatrix} 1 & \cos x \\ -\cos x & 1 \end{vmatrix} = 1 - \cos^2 x, \quad \begin{vmatrix} -\cos x & \cos x \\ -1 & 1 \end{vmatrix} = \cos x -\cos x = 0, \quad \begin{vmatrix} -\cos x & 1 \\ -1 & -\cos x \end{vmatrix} = \cos^2 x - 1. \]
Substituting these values back:
\[ \Delta = (1 - \cos^2 x) + 0 + (\cos^2 x - 1) = 2 - \sin^2 x.\]
Simplifying the expression:
\[\Delta = 2(2 - \sin^2 x).\]
This result corresponds to option (B).
The minimum value of \( \Delta \) occurs when \( \sin^2 x = 1 \), resulting in:
\[\Delta_{\min} = 2(2 - 1) = 2.\]
The maximum value of \( \Delta \) occurs when \( \sin^2 x = 0 \), resulting in:
\[\Delta_{\max} = 2(2 - 0) = 4.\]
Therefore, options (B), (C), and (D) are correct.
\(\boxed{(B), (C), \text{ and } (D) \text{ only}}\).