The solutions of the equation left|begin{arra | vedclass

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(D) Given the determinant equation: $\left|\begin{array}{ccc}1+\sin ^{2} x & \sin ^{2} x & \sin ^{2} x \ \cos ^{2} x & 1+\cos ^{2} x & \cos ^{2} x \

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

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(D) Given the determinant equation: $\left|\begin{array}{ccc}1+\sin ^{2} x & \sin ^{2} x & \sin ^{2} x \ \cos ^{2} x & 1+\cos ^{2} x & \cos ^{2} x \ 4 \sin 2 x & 4 \sin 2 x & 1+4 \sin 2 x\end{array}ight|=0$. Apply the row operation $R_{1} ightarrow R_{1} + R_{2} + R_{3}$: Since $\sin^{2} x + \cos^{2} x = 1$,the sum of each column in the first row becomes $1 + \sin^{2} x + \cos^{2} x + 4 \sin 2x = 2 + 4 \sin 2x$. Factoring out $(2 + 4 \sin 2x)$,we get: $(2 + 4 \sin 2x) \left|\begin{array}{ccc}1 & 1 & 1 \ \cos ^{2} x & 1+\cos ^{2} x & \cos ^{2} x \ 4 \sin 2 x & 4 \sin 2 x & 1+4 \sin 2 x \end{array}ight| = 0$. Applying $C_{2} ightarrow C_{2} - C_{1}$ and $C_{3} ightarrow C_{3} - C_{1}$: $(2 + 4 \sin 2x) \left|\begin{array}{ccc}1 & 0 & 0 \ \cos ^{2} x & 1 & 0 \ 4 \sin 2 x & 0 & 1 \end{array}ight| = 0$. This simplifies to $(2 + 4 \sin 2x)(1) = 0$,so $2 + 4 \sin 2x = 0$,which means $\sin 2x = -\frac{1}{2}$. For $0 The values of $2x$ where $\sin 2x = -\frac{1}{2}$ in the interval $(0, 2\pi)$ are $2x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$ and $2x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$. Thus,$x = \frac{7\pi}{12}$ and $x = \frac{11\pi}{12}$.

The solutions of the equation $\left|\begin{array}{ccc}1+\sin ^{2} x & \sin ^{2} x & \sin ^{2} x \\ \cos ^{2} x & 1+\cos ^{2} x & \cos ^{2} x \\ 4 \sin 2 x & 4 \sin 2 x & 1+4 \sin 2 x\end{array}\right|=0$ for $(0 < x < \pi)$ are:

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