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

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(A) Given the determinant equation: $\left|\begin{array}{lll} \sin x & \cos x & \cos x \ \cos x & \sin x & \cos x \ \cos x & \cos x & \sin x \end{ar

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(A) Given the determinant equation: $\left|\begin{array}{lll} \sin x & \cos x & \cos x \ \cos x & \sin x & \cos x \ \cos x & \cos x & \sin x \end{array}ight|=0$. Applying row operations $R_{1} ightarrow R_{1}-R_{2}$ and $R_{2} ightarrow R_{2}-R_{3}$,we get: $\left|\begin{array}{ccc} \sin x-\cos x & \cos x-\sin x & 0 \ 0 & \sin x-\cos x & \cos x-\sin x \ \cos x & \cos x & \sin x \end{array}ight|=0$. Taking $(\sin x-\cos x)$ common from $R_{1}$ and $R_{2}$: $(\sin x-\cos x)^{2} \left|\begin{array}{ccc} 1 & -1 & 0 \ 0 & 1 & -1 \ \cos x & \cos x & \sin x \end{array}ight|=0$. Expanding the determinant: $(\sin x-\cos x)^{2} [1(\sin x + \cos x) + 1(0 + \cos x)] = 0$. $(\sin x-\cos x)^{2} (\sin x + 2 \cos x) = 0$. This implies $\sin x = \cos x$ or $\sin x = -2 \cos x$. Case $1$: $	an x = 1 \implies x = \frac{\pi}{4}$. This value lies in the interval $[-\frac{\pi}{4}, \frac{\pi}{4}]$. Case $2$: $	an x = -2$. In the interval $[-\frac{\pi}{4}, \frac{\pi}{4}]$,$	an x$ ranges from $-1$ to $1$. Since $-2$ is outside this range,there is no solution for $	an x = -2$ in this interval. Thus,there is only $1$ distinct real root,which is $x = \frac{\pi}{4}$.

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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