The number of distinct real roots of the equa | vedclass

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(C) Let the determinant be $D$. Applying $R_1 	o R_1 - R_2$ and $R_2 	o R_2 - R_3$:$D = \begin{vmatrix} \cos x - \sin x & \sin x - \cos x & 0 \ 0 &

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$2$

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(C) Let the determinant be $D$. Applying $R_1 	o R_1 - R_2$ and $R_2 	o R_2 - R_3$:$D = \begin{vmatrix} \cos x - \sin x & \sin x - \cos x & 0 \ 0 & \cos x - \sin x & \sin x - \cos x \ \sin x & \sin x & \cos x \end{vmatrix} = 0$Taking $(\cos x - \sin x)$ common from $R_1$ and $R_2$:$D = (\cos x - \sin x)^2 \begin{vmatrix} 1 & -1 & 0 \ 0 & 1 & -1 \ \sin x & \sin x & \cos x \end{vmatrix} = 0$Expanding along $R_1$:$(\cos x - \sin x)^2 [1(\cos x + \sin x) - (-1)(0 + \sin x)] = 0$$(\cos x - \sin x)^2 [\cos x + \sin x + \sin x] = 0$$(\cos x - \sin x)^2 (\cos x + 2\sin x) = 0$This gives $\cos x - \sin x = 0$ or $\cos x + 2\sin x = 0$.Case $1$: $	an x = 1 \implies x = \frac{\pi}{4}$.Case $2$: $	an x = -\frac{1}{2} \implies x = \arctan(-\frac{1}{2})$.Since $\arctan(-\frac{1}{2}) \approx -0.46$ and $-\frac{\pi}{4} \approx -0.785$,both values lie in the interval $\left[ -\frac{\pi}{4}, \frac{\pi}{4} ight]$.Thus,there are $2$ distinct real roots.

The number of distinct real roots of the equation $\begin{vmatrix} \cos x & \sin x & \sin x \\ \sin x & \cos x & \sin x \\ \sin x & \sin x & \cos x \end{vmatrix} = 0$ in the interval $\left[ -\frac{\pi}{4}, \frac{\pi}{4} \right]$ is

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