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(C) Let the given determinant be $\Delta$. Adding $R_2$ and $R_3$ to $R_1$,we get: $\Delta = \left| {\begin{array}{*{20}{c}}{\sin x + 2\cos x}&{\sin x

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$1$

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(C) Let the given determinant be $\Delta$. Adding $R_2$ and $R_3$ to $R_1$,we get: $\Delta = \left| {\begin{array}{*{20}{c}}{\sin x + 2\cos x}&{\sin x + 2\cos x}&{\sin x + 2\cos x}\{\cos x}&{\sin x}&{\cos x}\{\cos x}&{\cos x}&{\sin x}\end{array}} ight| = 0$ Taking $(\sin x + 2\cos x)$ common from $R_1$: $\Delta = (\sin x + 2\cos x) \left| {\begin{array}{*{20}{c}}1&1&1\{\cos x}&{\sin x}&{\cos x}\{\cos x}&{\cos x}&{\sin x}\end{array}} ight| = 0$ Applying $C_2 	o C_2 - C_1$ and $C_3 	o C_3 - C_1$: $\Delta = (\sin x + 2\cos x) \left| {\begin{array}{*{20}{c}}1&0&0\{\cos x}&{\sin x - \cos x}&0\{\cos x}&0&{\sin x - \cos x}\end{array}} ight| = 0$ $\Delta = (\sin x + 2\cos x)(\sin x - \cos x)^2 = 0$ This implies $\sin x + 2\cos x = 0$ or $(\sin x - \cos x)^2 = 0$. Case $1$: $	an x = -2$. In the interval $[-\frac{\pi}{4}, \frac{\pi}{4}]$,$	an x$ ranges from $-1$ to $1$. Thus,$	an x = -2$ has no solution. Case $2$: $\sin x = \cos x \implies 	an x = 1$. In the interval $[-\frac{\pi}{4}, \frac{\pi}{4}]$,$	an x = 1$ at $x = \frac{\pi}{4}$. Thus,there is only $1$ distinct real root.

The number of distinct real roots of $\left| {\begin{array}{*{20}{c}}{\sin x}&{\cos x}&{\cos x}\\{\cos x}&{\sin x}&{\cos x}\\{\cos x}&{\cos x}&{\sin x}\end{array}} \right| = 0$ in the interval $-\frac{\pi}{4} \le x \le \frac{\pi}{4}$ is

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