1. Решить уравнения: a) cos(3x - π/3) = 1/2 б) 27(x^2 - 1) = 9(x + 2) в) (log₂x)² - 3log₂x + 2 = 0

1. Решение уравнений:

1. а) \( \cos(3x - \frac{\pi}{3}) = \frac{1}{2} \)
   \( 3x - \frac{\pi}{3} = \pm \frac{\pi}{3} + 2\pi k, k \in \mathbb{Z} \)
   1) \( 3x - \frac{\pi}{3} = \frac{\pi}{3} + 2\pi k \) \( \Rightarrow 3x = \frac{2\pi}{3} + 2\pi k \) \( \Rightarrow x = \frac{2\pi}{9} + \frac{2\pi k}{3} \)
   2) \( 3x - \frac{\pi}{3} = -\frac{\pi}{3} + 2\pi k \) \( \Rightarrow 3x = 2\pi k \) \( \Rightarrow x = \frac{2\pi k}{3} \)
   Ответ: \( x = \frac{2\pi k}{3} \) и \( x = \frac{2\pi}{9} + \frac{2\pi k}{3}, k \in \mathbb{Z} \).
2. б) \( 27(x^2 - 1) = 9(x + 2) \)
   \( 3(x^2 - 1) = x + 2 \)
   \( 3x^2 - 3 = x + 2 \)
   \( 3x^2 - x - 5 = 0 \)
   \( D = (-1)^2 - 4(3)(-5) = 1 + 60 = 61 \)
   \( x = \frac{1 \pm \sqrt{61}}{6} \)
   Ответ: \( x = \frac{1 \pm \sqrt{61}}{6} \).
3. в) \( (\log_2 x)^2 - 3\log_2 x + 2 = 0 \)
   Пусть \( y = \log_2 x \). Тогда \( y^2 - 3y + 2 = 0 \)
   \( (y-1)(y-2) = 0 \)
   \( y = 1 \) или \( y = 2 \)
   \( \log_2 x = 1 \) \( \Rightarrow x = 2^1 = 2 \)
   \( \log_2 x = 2 \) \( \Rightarrow x = 2^2 = 4 \)
   Ответ: \( x = 2, x = 4 \).