4. Решите уравнения: cos x = 1/6 sin (x - π/3) = 1 tg x = √3 ctg x = 2 5 cos²x - 8 cos x + 3 = 0

Решение:

1. \( \cos x = \frac{1}{6} \)
   \( x = \pm \arccos \left(\frac{1}{6}\right) + 2\pi n, n \in \mathbb{Z} \)
2. \( \sin \left(x - \frac{\pi}{3}\right) = 1 \)
   \( x - \frac{\pi}{3} = \frac{\pi}{2} + 2\pi k, k \in \mathbb{Z} \)
   \( x = \frac{\pi}{2} + \frac{\pi}{3} + 2\pi k \)
   \( x = \frac{5\pi}{6} + 2\pi k, k \in \mathbb{Z} \)
3. \( \operatorname{tg} x = \sqrt{3} \)
   \( x = \frac{\pi}{3} + \pi m, m \in \mathbb{Z} \)
4. \( \operatorname{ctg} x = 2 \)
   \( x = \operatorname{arcctg}(2) + \pi p, p \in \mathbb{Z} \)
5. \( 5\cos^2 x - 8\cos x + 3 = 0 \)
   Пусть \( y = \cos x \). Тогда \( 5y^2 - 8y + 3 = 0 \).
   \( D = (-8)^2 - 4 \cdot 5 \cdot 3 = 64 - 60 = 4 \)
   \( y_1 = \frac{8 + \sqrt{4}}{2 \cdot 5} = \frac{8+2}{10} = 1 \)
   \( y_2 = \frac{8 - \sqrt{4}}{2 \cdot 5} = \frac{8-2}{10} = \frac{6}{10} = \frac{3}{5} \)
   Случай 1: \( \cos x = 1 \)
   \( x = 2\pi n, n \in \mathbb{Z} \)
   Случай 2: \( \cos x = \frac{3}{5} \)
   \( x = \pm \arccos \left(\frac{3}{5}\right) + 2\pi q, q \in \mathbb{Z} \)

Ответ:
1. \( x = \pm \arccos \left(\frac{1}{6}\right) + 2\pi n, n \in \mathbb{Z} \)
2. \( x = \frac{5\pi}{6} + 2\pi k, k \in \mathbb{Z} \)
3. \( x = \frac{\pi}{3} + \pi m, m \in \mathbb{Z} \)
4. \( x = \operatorname{arcctg}(2) + \pi p, p \in \mathbb{Z} \)
5. \( x = 2\pi n \) и \( x = \pm \arccos \left(\frac{3}{5}\right) + 2\pi q, n, q \in \mathbb{Z} \).