1. Решить уравнения: a) 2 cos(2x - π/3) = √3 б) 5(x²-3) = 5(x+3) в) lg²x - 4lg x - 5 = 0

1. Решение уравнений:

1. \( 2 \cos(2x - \frac{\pi}{3}) = \sqrt{3} \)
   \( \cos(2x - \frac{\pi}{3}) = \frac{\sqrt{3}}{2} \)
   \( 2x - \frac{\pi}{3} = \pm \frac{\pi}{6} + 2\pi n, n \in \mathbb{Z} \)
   \( 2x = \frac{\pi}{3} \pm \frac{\pi}{6} + 2\pi n \)
   \( 2x_1 = \frac{\pi}{3} + \frac{\pi}{6} + 2\pi n = \frac{\pi}{2} + 2\pi n \Rightarrow x_1 = \frac{\pi}{4} + \pi n \)
   \( 2x_2 = \frac{\pi}{3} - \frac{\pi}{6} + 2\pi n = \frac{\pi}{6} + 2\pi n \Rightarrow x_2 = \frac{\pi}{12} + \pi n, n \in \mathbb{Z} \)
2. \( 5(x^2 - 3) = 5(x+3) \)
   \( x^2 - 3 = x + 3 \)
   \( x^2 - x - 6 = 0 \)
   \( D = (-1)^2 - 4(1)(-6) = 1 + 24 = 25 \)
   \( x = \frac{1 \pm \sqrt{25}}{2} = \frac{1 \pm 5}{2} \)
   \( x_1 = \frac{1+5}{2} = 3 \)
   \( x_2 = \frac{1-5}{2} = -2 \)
3. \( \lg^2 x - 4\lg x - 5 = 0 \)
   Пусть \( y = \lg x \). Тогда \( y^2 - 4y - 5 = 0 \)
   \( D = (-4)^2 - 4(1)(-5) = 16 + 20 = 36 \)
   \( y = \frac{4 \pm \sqrt{36}}{2} = \frac{4 \pm 6}{2} \)
   \( y_1 = \frac{4+6}{2} = 5 \)
   \( y_2 = \frac{4-6}{2} = -1 \)
   \( \lg x = 5 \Rightarrow x = 10^5 \)
   \( \lg x = -1 \Rightarrow x = 10^{-1} = \frac{1}{10} \)

Ответ: а) \( x = \frac{\pi}{4} + \pi n \) или \( x = \frac{\pi}{12} + \pi n, n \in \mathbb{Z} \); б) \( x = 3 \) или \( x = -2 \); в) \( x = 10^5 \) или \( x = 0.1 \).