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44) \( -3 \sin x \cos x<0 \)

Ask by Haynes Cox. in Italy
Mar 15,2025

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L'inequazione \(-3 \sin x \cos x < 0\) è vera quando \(\sin x \cos x > 0\). Questo accade: 1. Quando entrambi \(\sin x\) e \(\cos x\) sono positivi, nel primo quadrante: \(x \in \left(0, \frac{\pi}{2}\right)\) e periodi successivi. 2. Quando entrambi \(\sin x\) e \(\cos x\) sono negativi, nel terzo quadrante: \(x \in \left(\pi, \frac{3\pi}{2}\right)\) e periodi successivi. Quindi, la soluzione è: \[ x \in \left(2\pi k,\, 2\pi k + \frac{\pi}{2}\right) \cup \left(\pi + 2\pi k,\, \frac{3\pi}{2} + 2\pi k\right) \quad \text{per ogni } k \in \mathbb{Z}. \]

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To solve the inequality \( -3 \sin x \cos x < 0 \), we first simplify it to \( \sin x \cos x > 0 \) (since multiplying by -1 reverses the inequality). Knowing that \( \sin x \cos x = \frac{1}{2} \sin(2x) \), we deduce that our inequality reduces to \( \sin(2x) > 0 \). This means \( 2x \) must lie in intervals where the sine function is positive, which occurs in the ranges \( (0, \pi) \) and \( (2\pi, 3\pi) \). Thus, for the original \( x \), the solution converts to: - \( 0 < 2x < \pi \) leading to \( 0 < x < \frac{\pi}{2} \) - \( 2\pi < 2x < 3\pi \) giving \( \pi < x < \frac{3\pi}{2} \) So, \( x \) values satisfying \( -3 \sin x \cos x < 0 \) are in \( (0, \frac{\pi}{2}) \) and \( (\pi, \frac{3\pi}{2}) \).

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