Ответ:
Объяснение:
[tex]13. \; 2 \left(\dfrac{\sqrt3}{2} \sin 2x+ \dfrac12 \cos 2x\right) \leqslant 1\\\sin \dfrac{\pi}{3}\sin 2x+ \cos \dfrac{\pi}{3} \cos 2x \leqslant \dfrac12\\\cos \left(2x-\dfrac{\pi}{3}\right) \leqslant \dfrac12\\2x-\dfrac{\pi}{3} \in \left[ \dfrac{\pi}{3} +2\pi k; \dfrac{5\pi}{3}+2\pi k \right], k \in \mathbb{Z}\\2x \in \left[ \dfrac{2\pi}{3}+2\pi k; 2\pi+2\pi k\right]\\x \in \left[ \dfrac{\pi}{3}+\pi k; \pi+\pi k\right][/tex]
[tex]14. \; 3-4\cos^2x \leqslant 0\\2-4\cos^2x\leqslant -1\\\cos 2x \leqslant -\dfrac12\\2x \in \left[\dfrac{2\pi}{3}+2\pi k; \dfrac{4\pi}{3}+2\pi k\right], k \in \mathbb{Z}\\x \in \left[ \dfrac{\pi}{3}+\pi k; \dfrac{2\pi}{3}+\pi k \right]\\[/tex]
[tex]15. \; \dfrac{1}{\sin^2x}+\cot x-3 \leqslant 0\\1+\cot^2x+\cot x -3\leqslant 0\\\cot^2x+\cot x -2 \leqslant 0\\(\cot x+2)(\cot x -1)\leqslant 0\\\cot x \in [-2; 1]\\x\in \left[-arccot 2 +\pi k; \dfrac{\pi}{4}+\pi k \right], k \in \mathbb{Z}[/tex]
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Answers & Comments
Ответ:
Объяснение:
[tex]13. \; 2 \left(\dfrac{\sqrt3}{2} \sin 2x+ \dfrac12 \cos 2x\right) \leqslant 1\\\sin \dfrac{\pi}{3}\sin 2x+ \cos \dfrac{\pi}{3} \cos 2x \leqslant \dfrac12\\\cos \left(2x-\dfrac{\pi}{3}\right) \leqslant \dfrac12\\2x-\dfrac{\pi}{3} \in \left[ \dfrac{\pi}{3} +2\pi k; \dfrac{5\pi}{3}+2\pi k \right], k \in \mathbb{Z}\\2x \in \left[ \dfrac{2\pi}{3}+2\pi k; 2\pi+2\pi k\right]\\x \in \left[ \dfrac{\pi}{3}+\pi k; \pi+\pi k\right][/tex]
[tex]14. \; 3-4\cos^2x \leqslant 0\\2-4\cos^2x\leqslant -1\\\cos 2x \leqslant -\dfrac12\\2x \in \left[\dfrac{2\pi}{3}+2\pi k; \dfrac{4\pi}{3}+2\pi k\right], k \in \mathbb{Z}\\x \in \left[ \dfrac{\pi}{3}+\pi k; \dfrac{2\pi}{3}+\pi k \right]\\[/tex]
[tex]15. \; \dfrac{1}{\sin^2x}+\cot x-3 \leqslant 0\\1+\cot^2x+\cot x -3\leqslant 0\\\cot^2x+\cot x -2 \leqslant 0\\(\cot x+2)(\cot x -1)\leqslant 0\\\cot x \in [-2; 1]\\x\in \left[-arccot 2 +\pi k; \dfrac{\pi}{4}+\pi k \right], k \in \mathbb{Z}[/tex]