Задание
а) Реши уравнение:
| \(\log_{4}\) | \(\left(2^{2x} - 2\cos x - 4\sqrt{3}\sin^2 x\right) = x.\) |
|---|
- \(-\frac{2\pi}{3} + 2\pi n, n \in \mathbb{Z}\)
- \(\frac{2\pi}{3} + 2\pi n, n \in \mathbb{Z}\)
- \(-\frac{5\pi}{6} + 2\pi n, n \in \mathbb{Z}\)
- \(-\frac{\pi}{3} + 2\pi n, n \in \mathbb{Z}\)
- \(\frac{\pi}{6} + 2\pi n, n \in \mathbb{Z}\)
- \(-\frac{\pi}{6} + 2\pi n, n \in \mathbb{Z}\)
- \(\frac{5\pi}{6} + 2\pi n, n \in \mathbb{Z}\)
- \(\frac{\pi}{3} + 2\pi n, n \in \mathbb{Z}\)
б) Укажи все корни, принадлежащие отрезку \(\left[ \frac{13 \pi}{2} ; 8 \pi \right]\):
- \(\frac{22\pi}{3}\)
- \(\frac{41\pi}{6}\)
- другой ответ
- \(\frac{20\pi}{3}\)
- \(\frac{43\pi}{6}\)