1. Identify the matrix structure

The determinant is

$$\left| \begin{matrix} 1+\sin^2 x & \sin^2 x & \sin^2 x\\ \cos^2 x & 1+\cos^2 x & \cos^2 x\\ 4\sin 2x & 4\sin 2x & 1+4\sin 2x \end{matrix}\right|=0.$$

Let

$a=\sin^2 x,\qquad b=\cos^2 x,\qquad c=4\sin 2x.$

Then the matrix becomes

$$\begin{pmatrix} 1+a & a & a\\ b & 1+b & b\\ c & c & 1+c \end{pmatrix}.$$

2. Use column operations to simplify the determinant

Apply

$C_1 \to C_1-C_2,\qquad C_2 \to C_2-C_3.$

Then:

• First column becomes $\begin{pmatrix}(1+a)-a\\ b-(1+b)\\ c-c\end{pmatrix}=\begin{pmatrix}1\\ -1\\ 0\end{pmatrix}$
• Second column becomes $\begin{pmatrix}a-a\\ (1+b)-b\\ c-(1+c)\end{pmatrix}=\begin{pmatrix}0\\ 1\\ -1\end{pmatrix}$
• Third column remains $\begin{pmatrix}a\\ b\\ 1+c\end{pmatrix}$

So the determinant becomes

$$\left|\begin{matrix} 1 & 0 & a\\ -1 & 1 & b\\ 0 & -1 & 1+c \end{matrix}\right|.$$

3. Evaluate this determinant

Expand along the first row:

$$\Delta = 1\cdot \left|\begin{matrix}1 & b\\ -1 & 1+c\end{matrix}\right| + a\cdot \left|\begin{matrix}-1 & 1\\ 0 & -1\end{matrix}\right|.$$

Now,

$$\left|\begin{matrix}1 & b\\ -1 & 1+c\end{matrix}\right| = 1(1+c)-b(-1)=1+b+c,$$

and

$$\left|\begin{matrix}-1 & 1\\ 0 & -1\end{matrix}\right| = (-1)(-1)-0=1.$$

Hence

$$\Delta = (1+b+c)+a = 1+a+b+c.$$

Since

$a+b=\sin^2 x+\cos^2 x=1,$

we get

$$\Delta = 2+4\sin 2x.$$

Given $\Delta=0$,

$$2+4\sin 2x=0$$ $$\sin 2x=-\frac12.$$

4. Solve in the interval $0<x<\pi$

Since $0<x<\pi$, we have

$0<2x<2\pi.$

Now,

$$\sin 2x=-\frac12$$

in $(0,2\pi)$ gives

$$2x=\frac{7\pi}{6},\ \frac{11\pi}{6}.$$

Therefore,

$$x=\frac{7\pi}{12},\ \frac{11\pi}{12}.$$

5. Match with options

This corresponds to:

Option D:

$$\frac{7\pi}{12},\ \frac{11\pi}{12}.$$

6. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.