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Решите неравенство. \(\sqrt{2}tg 5x\cdot cos 5x\leq 1\)

Решение №32629: \( \left [\frac{3\pi}{20}+\frac{2\pi n}{5}; \frac{3\pi}{10}+\frac{2\pi n}{5}\right )\cup\left (\frac{3\pi}{10}+\frac{2\pi n}{5}; \frac{9\pi}{20}+\frac{2\pi n}{5}\right ], n \in \mathbb{Z}\)

Ответ: \( \left [\frac{3\pi}{20}+\frac{2\pi n}{5}; \frac{3\pi}{10}+\frac{2\pi n}{5}\right )\cup\left (\frac{3\pi}{10}+\frac{2\pi n}{5}; \frac{9\pi}{20}+\frac{2\pi n}{5}\right ], n \in \mathbb{Z}\)

Решите неравенство. \(2tg 3x\cdot cos 3x\leq \sqrt{3}\)

Решение №32630: \( \left [\frac{2\pi}{9}+\frac{2\pi n}{3}; \frac{\pi}{2}+\frac{2\pi n}{3}\right )\cup\left (\frac{\pi}{2}+\frac{2\pi n}{3}; \frac{7\pi}{9}+\frac{2\pi n}{3}\right ], n \in \mathbb{Z}\)

Ответ: \( \left [\frac{2\pi}{9}+\frac{2\pi n}{3}; \frac{\pi}{2}+\frac{2\pi n}{3}\right )\cup\left (\frac{\pi}{2}+\frac{2\pi n}{3}; \frac{7\pi}{9}+\frac{2\pi n}{3}\right ], n \in \mathbb{Z}\)

Решите неравенство. \(2tg 4x\cdot cos 4x\geq 1\)

Решение №32631: \( \left [\frac{\pi}{24}+\frac{\pi n}{2}; \frac{\pi}{8}+\frac{\pi n}{2}\right )\cup\left (\frac{\pi}{8}+\frac{\pi n}{2}; \frac{5\pi}{24}+\frac{\pi n}{2}\right ], n \in \mathbb{Z}\)

Ответ: \( \left [\frac{\pi}{24}+\frac{\pi n}{2}; \frac{\pi}{8}+\frac{\pi n}{2}\right )\cup\left (\frac{\pi}{8}+\frac{\pi n}{2}; \frac{5\pi}{24}+\frac{\pi n}{2}\right ], n \in \mathbb{Z}\)

Решите неравенство. \(\sqrt{2}tg 6x\cdot cos 6x\geq 1\)

Решение №32632: \( \left [\frac{\pi}{24}+\frac{\pi n}{3}; \frac{\pi}{12}+\frac{\pi n}{3}\right )\cup\left (\frac{\pi}{12}+\frac{\pi n}{3}; \frac{\pi}{8}+\frac{\pi n}{3}\right ], n \in \mathbb{Z}\)

Ответ: \( \left [\frac{\pi}{24}+\frac{\pi n}{3}; \frac{\pi}{12}+\frac{\pi n}{3}\right )\cup\left (\frac{\pi}{12}+\frac{\pi n}{3}; \frac{\pi}{8}+\frac{\pi n}{3}\right ], n \in \mathbb{Z}\)

Решите неравенство. \(2ctg 8x\cdot sin 8x+\sqrt{3}\leq 0\)

Решение №32633: \( \left [\frac{5\pi}{48}+\frac{\pi n}{4}; \frac{\pi}{8}+\frac{\pi n}{4}\right )\cup\left (\frac{\pi}{8}+\frac{\pi n}{4}; \frac{7\pi}{48}+\frac{\pi n}{4}\right ], n \in \mathbb{Z}\)

Ответ: \( \left [\frac{5\pi}{48}+\frac{\pi n}{4}; \frac{\pi}{8}+\frac{\pi n}{4}\right )\cup\left (\frac{\pi}{8}+\frac{\pi n}{4}; \frac{7\pi}{48}+\frac{\pi n}{4}\right ], n \in \mathbb{Z}\)

Решите неравенство. \(2ctg 5x\cdot sin 5x+\sqrt{2}\leq 0\)

Решение №32634: \( \left [\frac{3\pi}{20}+\frac{2\pi n}{5}; \frac{\pi}{5}+\frac{2\pi n}{5}\right )\cup\left (\frac{\pi}{5}+\frac{2\pi n}{5}; \frac{\pi}{4}+\frac{2\pi n}{5}\right ], n \in \mathbb{Z}\)

Ответ: \( \left [\frac{3\pi}{20}+\frac{2\pi n}{5}; \frac{\pi}{5}+\frac{2\pi n}{5}\right )\cup\left (\frac{\pi}{5}+\frac{2\pi n}{5}; \frac{\pi}{4}+\frac{2\pi n}{5}\right ], n \in \mathbb{Z}\)

Решите неравенство. \(2ctg 10x\cdot sin 10x\geq \sqrt{2}\)

Решение №32635: \( \left [-\frac{\pi}{40}+\frac{\pi n}{5}; \frac{\pi n}{5}\right )\cup\left (\frac{\pi n}{5}; \frac{\pi}{40}+\frac{\pi n}{5}\right ], n \in \mathbb{Z}\)

Ответ: \( \left [-\frac{\pi}{40}+\frac{\pi n}{5}; \frac{\pi n}{5}\right )\cup\left (\frac{\pi n}{5}; \frac{\pi}{40}+\frac{\pi n}{5}\right ], n \in \mathbb{Z}\)

Решите неравенство. \(2ctg 12x\cdot sin 12x\geq \sqrt{3}\)

Решение №32636: \( \left [-\frac{\pi}{72}+\frac{\pi n}{6}; \frac{\pi n}{6}\right )\cup\left (\frac{\pi n}{6}; \frac{\pi}{72}+\frac{\pi n}{6}\right ], n \in \mathbb{Z}\)

Ответ: \( \left [-\frac{\pi}{72}+\frac{\pi n}{6}; \frac{\pi n}{6}\right )\cup\left (\frac{\pi n}{6}; \frac{\pi}{72}+\frac{\pi n}{6}\right ], n \in \mathbb{Z}\)

Решите неравенство. \(\frac{sin 8x}{sin 4x}\leq \sqrt{2}\)

Решение №32637: \( \left [\frac{\pi}{16}+\frac{\pi n}{2}; \frac{\pi}{4}+\frac{\pi n}{2}\right )\cup\left (\frac{\pi}{4}+\frac{\pi n}{2}; \frac{7\pi}{16}+\frac{\pi n}{2}\right ], n \in \mathbb{Z}\)

Ответ: \( \left [\frac{\pi}{16}+\frac{\pi n}{2}; \frac{\pi}{4}+\frac{\pi n}{2}\right )\cup\left (\frac{\pi}{4}+\frac{\pi n}{2}; \frac{7\pi}{16}+\frac{\pi n}{2}\right ], n \in \mathbb{Z}\)

Решите неравенство. \(\frac{sin 4x}{sin 2x}\leq \sqrt{3}\)

Решение №32638: \( \left [-\frac{\pi}{12}+\pi n; \pi n\right )\cup\left (\pi n; \frac{\pi}{12}+\pi n\right ], n \in \mathbb{Z}\)

Ответ: \( \left [-\frac{\pi}{12}+\pi n; \pi n\right )\cup\left (\pi n; \frac{\pi}{12}+\pi n\right ], n \in \mathbb{Z}\)