Question

дам 50 баллов! срочно! задание на фото1

Answer 1

$\sqrt{3}\cdot tg\left (3x+\dfrac{\pi}{3}\right ) - 1 = 0\\\\\\\sqrt{3}tg\left (3x+\dfrac{\pi}{3}\right ) = 1\\\\\\tg\left (3x+\dfrac{\pi}{3}\right ) = \dfrac{1}{\sqrt{3}}\\\\\\tg\left (3x+\dfrac{\pi}{3}\right ) = \dfrac{\sqrt{3}}{3}\\\\\\3x + \dfrac{\pi}{3} = \dfrac{\pi}{6} + \pi k\\\\\\3x = -\dfrac{\pi}{6} + \pi k\\\\\\\boxed{x = -\dfrac{\pi}{18} + \dfrac{\pi k}{3}}\ ,\ k\in\mathbb{Z}$