The system of equations has a non-trivial solution if and only if $\begin{vmatrix} \sin 3\theta & -2 & 3 \\ \cos 2\theta & 8 & -7 \\ 2 & 14 & 11 \end{vmatrix} = 0$
Step-by-Step Solution
Key Concept: For a homogeneous system to have non-trivial solutions, the coefficient matrix determinant must equal zero.
Applying $R_2 \to R_2 + 4R_1$ and $R_3 \to R_3 + 7R_1$, the determinant becomes $\begin{vmatrix} \sin 3\theta & -2 & 3 \\ \cos 2\theta + 4\sin 3\theta & 0 & 5 \\ 2 + 7\sin 3\theta & 0 & 10 \end{vmatrix}$. Expanding along $C_2$: $2(\cos 2\theta + 4\sin 3\theta) - (2 + 7\sin 3\theta) = 0$, which gives $2\cos 2\theta - \sin 3\theta = 0$. This simplifies to $4\sin^2\theta - (3\sin\theta - 4\sin^3\theta) = 0$, leading to $\sin\theta(4\sin\theta + 4\sin\theta - 3) = 0$, so $\sin\theta = 0$ or $\sin\theta = 1/2$. Since $\sin\theta = -3/2$ is impossible, we have $\theta = n\pi$ or $\theta = n\pi + (-1)^n\pi/6$.
Correct Answer: 1