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(C) For the system of linear equations to have a non-trivial solution, the determinant of the coefficient matrix must be zero. The system is: $4 \sin

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$\sin^{-1}\left(\frac{3}{16}\right)$

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(C) For the system of linear equations to have a non-trivial solution, the determinant of the coefficient matrix must be zero. The system is: $4 \sin 	heta x - 3y + z = 0$ $x - 6 \cos 2	heta y + z = 0$ $3x - 12y + 4z = 0$ The determinant is: $\Delta = \begin{vmatrix} 4 \sin 	heta & -3 & 1 \ 1 & -6 \cos 2	heta & 1 \ 3 & -12 & 4 \end{vmatrix} = 0$ Applying row operations $R_1 ightarrow 4R_1 - R_3$: $\begin{vmatrix} 16 \sin 	heta - 3 & 0 & 0 \ 1 & -6 \cos 2	heta & 1 \ 3 & -12 & 4 \end{vmatrix} = 0$ Expanding along the first row: $(16 \sin 	heta - 3) [(-6 \cos 2	heta)(4) - (1)(-12)] = 0$ $(16 \sin 	heta - 3) [-24 \cos 2	heta + 12] = 0$ $12(16 \sin 	heta - 3)(1 - 2 \cos 2	heta) = 0$ This gives $16 \sin 	heta = 3$ or $2 \cos 2	heta = 1$. $16 \sin 	heta = 3 \implies \sin 	heta = \frac{3}{16} \implies 	heta = \sin^{-1}\left(\frac{3}{16}ight)$. $2 \cos 2	heta = 1 \implies \cos 2	heta = \frac{1}{2} \implies 2	heta = \frac{\pi}{3} \implies 	heta = \frac{\pi}{6}$.

$A$ value of $\theta$ for which the following system of equations has a non-trivial solution is:
$(4 \sin \theta) x - 3y + z = 0$
$x - (6 \cos 2\theta) y + z = 0$
$3x - 12y + 4z = 0$

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$A$ value of $\theta$ for which the following system of equations has a non-trivial solution is: $(4 \sin \theta) x - 3y + z = 0$ $x - (6 \cos 2\theta) y + z = 0$ $3x - 12y + 4z = 0$