1. Statement of the problem: Simplify the expression $\sin^2\theta - 2\cos\theta + \tfrac{1}{4}$.\n2. Use the Pythagorean identity $\sin^2\theta = 1 - \cos^2\theta$ to rewrite the expression in terms of $\cos\theta$.\n3. Substitute the identity and combine like terms: $$\sin^2\theta - 2\cos\theta + \tfrac{1}{4} = (1 - \cos^2\theta) - 2\cos\theta + \tfrac{1}{4} = -\cos^2\theta - 2\cos\theta + \tfrac{5}{4}.\n$$\n4. Complete the square for the $\cos\theta$ terms to factor the quadratic: $$-\cos^2\theta - 2\cos\theta + \tfrac{5}{4} = -\bigl(\cos^2\theta + 2\cos\theta\bigr) + \tfrac{5}{4} = -\bigl(\cos\theta + 1\bigr)^2 + 1 + \tfrac{5}{4} = -\bigl(\cos\theta + 1\bigr)^2 + \tfrac{9}{4}.\n$$\n5. Recognize the result as a difference of squares and factor: $$\tfrac{9}{4} - \bigl(\cos\theta + 1\bigr)^2 = \bigl(\tfrac{3}{2} - (\cos\theta + 1)\bigr)\bigl(\tfrac{3}{2} + (\cos\theta + 1)\bigr) = \bigl(\tfrac{1}{2} - \cos\theta\bigr)\bigl(\tfrac{5}{2} + \cos\theta\bigr).\n$$\n6. Final answer: $$\sin^2\theta - 2\cos\theta + \tfrac{1}{4} = \bigl(\tfrac{1}{2} - \cos\theta\bigr)\bigl(\tfrac{5}{2} + \cos\theta\bigr).\n