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Home » $ 7a + 3b + c = -6 $ \quad (5) - wispro

$ 7a + 3b + c = -6 $ \quad (5) - wispro

📅 March 11, 2026 👤 scraface

Mar 11, 2026

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📌 \boxed{\dfrac{12}{7}}Question: Find all angles $ \theta \in [0^\circ, 360^\circ] $ that satisfy $ 2\cos(2\theta) + 1 = 0 $, inspired by periodic patterns in environmental data trends.

📌 Solution: Start by solving $ 2\cos(2\theta) + 1 = 0 \Rightarrow \cos(2\theta) = -\frac{1}{2} $. The general solutions for $ 2\theta $ are $ 2\theta = 120^\circ + 360^\circ k $ or $ 2\theta = 240^\circ + 360^\circ k $, where $ k \in \mathbb{Z} $. Dividing by 2, $ \theta = 60^\circ + 180^\circ k $ or $ \theta = 120^\circ + 180^\circ k $. For $ \theta \in [0^\circ, 360^\circ] $, the solutions are $ 60^\circ, 120^\circ, 240^\circ, 300^\circ $.

📌 \boxed{60^\circ, 120^\circ, 240^\circ, 300^\circ}

📌 Question: Let $ z^4 + z^2 + 1 = 0 $. Express the maximum imaginary part of a root as $ \sin \theta $, where $ \theta \in [0^\circ, 360^\circ] $, reflecting harmonic analysis in linguistic frequency models.

📌 Solution: Factor $ z^4 + z^2 + 1 = (z^2 + z + 1)(z^2 - z + 1) = 0 $. Solving $ z^2 + z + 1 = 0 $ gives roots $ z = \frac{-1 \pm i\sqrt{3}}{2} $, with imaginary parts $ \pm \frac{\sqrt{3}}{2} $. Solving $ z^2 - z + 1 = 0 $ gives $ z = \frac{1 \pm i\sqrt{3}}{2} $, with imaginary parts $ \pm \frac{\sqrt{3}}{2} $. The maximum imaginary part is $ \frac{\sqrt{3}}{2} = \sin 60^\circ $.

📌 \boxed{60^\circ}

📌 Question: If $ \mathbf{a}, \mathbf{b}, \mathbf{c} $ are unit vectors in $ \mathbb{R}^3 $ with $ \mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{c} = \mathbf{c} \cdot \mathbf{a} = \frac{1}{2} $, find the maximum value of $ \|\mathbf{a} + \mathbf{b} + \mathbf{c}\| $, modeling directional consistency in linguistic feature spaces.

📌 Solution: Compute $ \|\mathbf{a} + \mathbf{b} + \mathbf{c}\|^2 = 3 + 2(\mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} + \mathbf{c} \cdot \mathbf{a}) = 3 + 2\left(\frac{1}{2} + \frac{1}{2} + \frac{1}{2}\right) = 3 + 3 = 6 $. Thus, the norm is $ \sqrt{6} $.

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