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david muir muir - NBX Soluciones

📅 May 3, 2026 👤 scraface

May 03, 2026

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📌 2\sin z + \sqrt{3} = 0 \implies \sin z = -\frac{\sqrt{3}}{2}

📌 الزوايا $z$ التي تحقق $\sin z = -\frac{\sqrt{3}}{2}$ تقع في الربعين الثالث والرابع. الزاوية المرجعية هي $60^\circ$ لأن $\sin 60^\circ = \frac{\sqrt{3}}{2}$. لذلك، الحلول في الفترة $[0^\circ, 360^\circ]$ هي:

📌 z = 180^\circ + 60^\circ = 240^\circ \quad \text{و} \quad z = 360^\circ - 60^\circ = 300^\circ

📌 \boxed{240^\circ} \quad \text{و} \quad \boxed{300^\circ}

📌 الإجابة النهائية:** الزوايا هي $\boxed{240^\circ}$ و $\boxed{300^\circ}$.

📌 السؤال: أوجد المتجه $\mathbf{v}$ بحيث أن $\mathbf{v} \times \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 3 \\ -2 \end{pmatrix}$.

📌 الحل: ليكن $\mathbf{v} = \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}$. حاصل الضرب التقاطعي هو:

📌 \mathbf{v} \times \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \cdot 0 - v_3 \cdot 0 \\ v_3 \cdot 1 - v_1 \cdot 0 \\ v_1 \cdot 0 - v_2 \cdot 1 \end{pmatrix} = \begin{pmatrix} 0 \\ v_3 \\ -v_2 \end{pmatrix}

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