Question:

<p><strong>Step 1: Simplify the denominator.</strong></p><p>$\sin(x - \pi) = -\sin(x)$</p><p>So the denominator is $(\cot 2x - \cot x)(-\sin x)$</p><p><strong>Step 2: Simplify the cotangent difference.</strong></p><p>$\cot 2x - \cot x = \frac{\cos 2x}{\sin 2x} - \frac{\cos x}{\sin x}$</p><p>$= \frac{\cos 2x \sin x - \cos x \sin 2x}{\sin 2x \sin x} = \frac{-\sin(2x - x)}{\sin 2x \sin x} = \frac{-\sin x}{\sin 2x \sin x}$</p><p>$= \frac{-1}{\sin 2x}$ (when $\sin x \neq 0$)</p><p><strong>Step 3: Rewrite the complete denominator.</strong></p><p>Denominator $= \frac{-1}{\sin 2x} \cdot (-\sin x) = \frac{\sin x}{\sin 2x}$</p><p><strong>Step 4: Set up the equation for the fraction to equal zero.</strong></p><p>$\frac{\cos 2x - 3\cos x + 1}{\frac{\sin x}{\sin 2x}} = 0$</p><p>This requires: $\cos 2x - 3\cos x + 1 = 0$ AND the denominator $\neq 0$</p><p><strong>Step 5: Solve the numerator equation.</strong></p><p>Using $\cos 2x = 2\cos^2 x - 1$:</p><p>$2\cos^2 x - 1 - 3\cos x + 1 = 0$</p><p>$2\cos^2 x - 3\cos x = 0$</p><p>$\cos x(2\cos x - 3) = 0$</p><p>So $\cos x = 0$ or $\cos x = \frac{3}{2}$ (impossible)</p><p><strong>Step 6: Check if $\cos x = 0$ is valid.</strong></p><p>If $\cos x = 0$, then $x = \frac{\pi}{2} + n\pi$, so $\sin x = \pm 1 \neq 0$</p><p>But we need $\sin x \neq 0$ for Step 2. However, we also need denominator $= \frac{\sin x}{\sin 2x} \neq 0$</p><p>Since $\sin 2x = 2\sin x \cos x = 0$ when $\cos x = 0$, the denominator becomes $\frac{\sin x}{0}$ which is undefined.</p><p><strong>Step 7: Conclusion.</strong></p><p>The only potential solution $\cos x = 0$ makes the denominator undefined. Therefore, there is no value of $x$ for which the equation holds.</p><p><strong>∴ Answer: D</strong></p>