Question:

<p><strong>Step 1:</strong> For non-trivial solutions, the coefficient matrix determinant must be zero:</p><p>$$\begin{vmatrix} 1 & 3 & 7 \\ -1 & 4 & 7 \\ \sin 3\theta & \cos 2\theta & 2 \end{vmatrix} = 0$$</p><p><strong>Step 2:</strong> Expand along the first row:</p><p>$$1(4 \cdot 2 - 7\cos 2\theta) - 3(-1 \cdot 2 - 7\sin 3\theta) + 7(-\cos 2\theta - 4\sin 3\theta) = 0$$</p><p>$$8 - 7\cos 2\theta + 6 + 21\sin 3\theta - 7\cos 2\theta - 28\sin 3\theta = 0$$</p><p>$$14 - 14\cos 2\theta - 7\sin 3\theta = 0$$</p><p>$$2 - 2\cos 2\theta - \sin 3\theta = 0$$</p><p><strong>Step 3:</strong> Rearrange: $$\sin 3\theta = 2(1 - \cos 2\theta) = 2 \cdot 2\sin^2\theta = 4\sin^2\theta$$</p><p><strong>Step 4:</strong> Use $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$:</p><p>$$3\sin\theta - 4\sin^3\theta = 4\sin^2\theta$$</p><p>$$\sin\theta(3 - 4\sin^2\theta - 4\sin\theta) = 0$$</p><p><strong>Step 5:</strong> Either $\sin\theta = 0$ (giving $\theta = \pi$, excluded) or solve $4\sin^2\theta + 4\sin\theta - 3 = 0$</p><p>Using quadratic formula: $\sin\theta = \frac{-4 \pm \sqrt{16+48}}{8} = \frac{-4 \pm 8}{8}$</p><p>So $\sin\theta = \frac{1}{2}$ or $\sin\theta = -\frac{3}{2}$ (invalid)</p><p><strong>Step 6:</strong> For $\sin\theta = \frac{1}{2}$ in $(0,\pi)$: $\theta = \frac{\pi}{6}, \frac{5\pi}{6}$</p><p>∴ Answer: <strong>2</strong></p>