The number of distinct real roots of left|beg | vedclass

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(C) Given the determinant equation: $\left|\begin{array}{ccc}\sin x & \cos x & \cos x \ \cos x & \sin x & \cos x \ \cos x & \cos x & \sin x\end{arra

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$1$

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(C) Given the determinant equation: $\left|\begin{array}{ccc}\sin x & \cos x & \cos x \ \cos x & \sin x & \cos x \ \cos x & \cos x & \sin x\end{array}ight|=0$. Applying row operations $R_1 	o R_1 + R_2 + R_3$, we get: $\left|\begin{array}{ccc}\sin x + 2\cos x & \sin x + 2\cos x & \sin x + 2\cos x \ \cos x & \sin x & \cos x \ \cos x & \cos x & \sin x\end{array}ight|=0$. Taking $(\sin x + 2\cos x)$ common from $R_1$: $(\sin x + 2\cos x) \left|\begin{array}{ccc}1 & 1 & 1 \ \cos x & \sin x & \cos x \ \cos x & \cos x & \sin x\end{array}ight|=0$. Applying $C_2 	o C_2 - C_1$ and $C_3 	o C_3 - C_1$: $(\sin x + 2\cos x) \left|\begin{array}{ccc}1 & 0 & 0 \ \cos x & \sin x - \cos x & 0 \ \cos x & 0 & \sin x - \cos x\end{array}ight|=0$. Expanding along $R_1$: $(\sin x + 2\cos x)(\sin x - \cos x)^2 = 0$. This gives $	an x = 1$ or $	an x = -2$. For $x \in [-\frac{\pi}{4}, \frac{\pi}{4}]$, $	an x = 1$ gives $x = \frac{\pi}{4}$. Since $	an x = -2$ is not possible in this interval (as $	an x \in [-1, 1]$), the only solution is $x = \frac{\pi}{4}$. Thus, the number of distinct real roots is $1$.

The number of distinct real roots of $\left|\begin{array}{lll}\sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{array}\right|=0$ in the interval $-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$ is

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