Question

даю 75 баллов за выполнение задания, срочно!

Answer 1

$\sqrt{3}*tg(3x+\frac{\pi }{3})-1=0\\\\\sqrt{3}*tg(3x+\frac{\pi }{3})=1\\\\tg(3x+\frac{\pi }{3})=\frac{1}{\sqrt{3}}\\\\3x+\frac{\pi }{3}=arctg\frac{1}{\sqrt{3}} +\pi n ,n\in Z\\\\3x+\frac{\pi }{3}=\frac{\pi }{6}+\pi n,n\in Z \\\\3x=\frac{\pi }{6}-\frac{\pi }{3} +\pi n,n\in Z\\\\3x=-\frac{\pi }{6} +\pi n,n\in Z\\\\\boxed{x=-\frac{\pi }{18}+\frac{\pi n }{3},n\in Z}$