<p>If the quadratic equation \((\log_2(\sin\theta))x^2 + 2x - 1 = 0\) has integral roots, then \(\pi/\theta\) can be:</p>
Step-by-Step Solution
Key Concept: For a quadratic with integral roots, use Vieta's formulas to constrain the coefficient $\log_2(\sin\theta)$, then recognize that $\sin\theta$ must equal a power of 2 within $[-1,1]$.
<p><strong>Step 1:</strong> Let the integral roots be $r$ and $s$. By Vieta's formulas:</p><ul><li>Sum: $r + s = -\frac{2}{\log_2(\sin\theta)}$</li><li>Product: $rs = -\frac{1}{\log_2(\sin\theta)}$</li></ul><p><strong>Step 2:</strong> From the product formula: $\log_2(\sin\theta) = -\frac{1}{rs}$ where $rs$ is an integer product.</p><p><strong>Step 3:</strong> For $r + s$ to be an integer: $r + s = -\frac{2}{-1/rs} = 2rs$ ✓ (automatically satisfied)</p><p><strong>Step 4:</strong> Since $\sin\theta \in (0,1]$ for the logarithm to exist, we need $\log_2(\sin\theta) \leq 0$. Thus $rs > 0$ (both roots same sign).</p><p><strong>Step 5:</strong> Testing cases: If $rs = -1$, then $\log_2(\sin\theta) = 1 \Rightarrow \sin\theta = 2$ (impossible). If $rs = 1$, then $\log_2(\sin\theta) = -1 \Rightarrow \sin\theta = 1/2$ ✓</p><p><strong>Step 6:</strong> With $\sin\theta = 1/2$: roots satisfy $r + s = -2$ and $rs = 1$, giving $r = s = -1$. The equation becomes $-x^2 + 2x - 1 = 0$ or $(x+1)^2 = 0$ ✓</p><p><strong>Step 7:</strong> $\sin\theta = 1/2 \Rightarrow \theta = \frac{\pi}{6} + 2\pi k$ or $\theta = \frac{5\pi}{6} + 2\pi k$</p><p>∴ $\frac{\pi}{\theta}$ can be: $\frac{1}{6}, \frac{1}{5\cdot 6}, \frac{1}{6 + 2k}, ...$</p><p>Common values: <strong>$\frac{1}{6}, \frac{6}{13}, \frac{6}{25}, \frac{6}{37}$</strong> (for $k=0,1,2,3$)</p>
Correct Answer: A,B,C,D