If $f(x) = \begin{vmatrix} \cos(2x) & \cos(2x) & \sin(2x) \\ -\cos x & \cos x & -\sin x \\ \sin x & \sin x & \cos x \end{vmatrix}$, then

$\text{Given: } f(x) = \begin{vmatrix} \cos 2x & \cos 2x & \sin 2x \\ -\cos x & \cos x & -\sin x \\ \sin x & \sin x & \cos x \end{vmatrix}$ $\text{To find: which of (1)--(4) are TRUE}$ $1.\ \text{expand the determinant} \Rightarrow f(x)$ $2.\ \text{maximum/minimum of } f$ $3.\ \text{zeros of } f' \Rightarrow \text{decide}$ $f(x) = \cos 2x\begin{vmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{vmatrix} - \cos 2x\begin{vmatrix} -\cos x & -\sin x \\ \sin x & \cos x \end{vmatrix} + \sin 2x\begin{vmatrix} -\cos x & \cos x \\ \sin x & \sin x \end{vmatrix}$ $= \cos 2x\,(1) - \cos 2x\,(-\cos 2x) + \sin 2x\,(-\sin 2x)$ $= \cos 2x + \cos^2 2x - \sin^2 2x$ $f(x) = \cos 2x + \cos 4x = 2\cos x\cos 3x$ $f(0) = 1 + 1 = 2 = \text{global maximum} \Rightarrow \text{(3) TRUE},\ \text{(4) FALSE}$ $f'(x) = -2\sin 2x - 4\sin 4x$ $f'(x) = -2\sin 2x\,(1 + 4\cos 2x)$ $\sin 2x = 0 \Rightarrow x = \dfrac{k\pi}{2} = -\dfrac{\pi}{2},\ 0,\ \dfrac{\pi}{2} \quad (3 \text{ points})$ $\cos 2x = -\dfrac14 \Rightarrow x = \pm 0.912,\ \pm 2.230 \quad (4 \text{ points})$ $f' = 0 \text{ at } 3 + 4 = 7 \text{ points} \Rightarrow \text{(2) TRUE},\ \text{(1) FALSE}$ $\boxed{\text{Answer: (2) and (3)}}$