1. Compute the determinant

Given

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

Let

$$\quad \sin 2x=2sc.$$

So

$$f(x)=\begin{vmatrix} \cos 2x & \cos 2x & \sin 2x\\ -c & c & -s\\ s & s & c \end{vmatrix}.$$

Expand along the first row:

$$f(x)=\cos 2x\begin{vmatrix}c & -s\\ s & c\end{vmatrix} -\cos 2x\begin{vmatrix}-c & -s\\ s & c\end{vmatrix} +\sin 2x\begin{vmatrix}-c & c\\ s & s\end{vmatrix}.$$

Now compute each minor:

• First minor:

$$\begin{vmatrix}c & -s\\ s & c\end{vmatrix}=c^2+s^2=1$$

• Second minor:

$$\begin{vmatrix}-c & -s\\ s & c\end{vmatrix}=(-c)(c)-(-s)(s)=-c^2+s^2=-(c^2-s^2)=-\cos 2x$$

• Third minor:

$$\begin{vmatrix}-c & c\\ s & s\end{vmatrix}=(-c)(s)-c(s)=-2cs=-\sin 2x$$

Hence

$$f(x)=\cos 2x(1)-\cos 2x(-\cos 2x)+\sin 2x(-\sin 2x).$$

So

$$f(x)=\cos 2x+\cos^2 2x-\sin^2 2x.$$

Using $\cos^2\theta-\sin^2\theta=\cos 2\theta,$ we get

$$f(x)=\cos 2x+\cos 4x.$$

2. Simplify the function

Use

$$\cos 4x=2\cos^2 2x-1.$$

Then

$$f(x)=\cos 2x+2\cos^2 2x-1.$$

Let

$$$$

Then

$$f(x)=2t^2+t-1.$$

This is an upward-opening quadratic in $t$.

3. Find maximum and minimum values

For $g(t)=2t^2+t-1,$ its vertex is at

$$t=-\frac{b}{2a}=-\frac{1}{4},$$

which lies in $[-1,1]$. So minimum occurs at $t=-\frac14$:

$$f_{\min}=2\left(\frac{1}{16}\right)-\frac14-1=\frac18-\frac14-1=-\frac98.$$

Maximum on $[-1,1]$ occurs at an endpoint:

• At $t=1$:

$$f=2(1)^2+1-1=2$$

• At $t=-1$:

$$f=2(1)-1-1=0$$

So the maximum value is $2$, attained when

$$\cos 2x=1 \iff 2x=2n\pi \iff x=n\pi.$$

In particular, at $x=0$, $f(0)=2,$ which is the maximum.

Therefore:

• A is false
• B is true

4. Find critical points of $f$ in $(-\pi,\pi)$

Differentiate:

$$f(x)=\cos 2x+\cos 4x$$

so

$$f'(x)=-2\sin 2x-4\sin 4x.$$

Using $\sin 4x=2\sin 2x\cos 2x,$ we get

$$f'(x)=-2\sin 2x-8\sin 2x\cos 2x$$ $$f'(x)=-2\sin 2x(1+4\cos 2x).$$

Thus

$$f'(x)=0$$

when either

Case 1: $\sin 2x=0$

$$2x=n\pi \iff x=\frac{n\pi}{2}.$$

In $(-\pi,\pi)$, these are

$$-\frac{\pi}{2},\ 0,\ \frac{\pi}{2}.$$

Also $x=\pm\pi$ are excluded.

So this gives 3 points.

Case 2: $1+4\cos 2x=0$

$$\cos 2x=-\frac14.$$

Let

$$\alpha=\cos^{-1}\left(-\frac14\right), \quad \alpha\in(0,\pi).$$

Then

$$2x=\pm \alpha+2k\pi.$$

Over $2x\in(-2\pi,2\pi)$, this gives four distinct solutions, hence in $x\in(-\pi,\pi)$ there are 4 points.

So total number of points where $f'(x)=0$ in $(-\pi,\pi)$ is

$$3+4=7,$$

which is more than three.

Therefore:

• C is true
• D is false

5. Final selection

The correct options are:

$$\boxed{B,\ C}$$

6. Comparison with stored correct answer

Stored correct answer: B, C

This matches our derived answer exactly.