Question

1.tg(6x-pi/3)=корень 3
2.ctg(x-pi/5)= 1​

Answer 1

1)

$tg\left (6x - \dfrac{\pi}{3}\right ) = \sqrt{3}\\\\\\6x - \dfrac{\pi}{3} = arctg(\sqrt{3}) + \pi k\\\\\\6x = \dfrac{\pi}{3} + \pi k + \dfrac{\pi}{3}\\\\\\6x = \dfrac{2\pi}{3} + \pi k\\\\\\\boxed{x = \dfrac{\pi}{9} + \dfrac{\pi k}{6}}\ ,\ k\in\mathbb{Z}$

2)

$ctg\left (x-\dfrac{\pi}{5}\right ) = 1\\\\\\x - \dfrac{\pi}{5} = arcctg1 + \pi k\\\\\\x = \dfrac{\pi}{4} + \pi k + \dfrac{\pi}{5}\\\\\\\boxed{x = \dfrac{9\pi}{20} + \pi k}\ ,\ k\in\mathbb{Z}$