Question

Хелп 4)
Срочнооо
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Answer 1

$tg(6x-\frac{\pi }{3})=-\sqrt{3}\\\\6x-\frac{\pi }{3}=arctg(-\sqrt{3})+\pi n,n\in z\\\\6x-\frac{\pi }{3}=-\frac{\pi }{3}+\pi n,n\in z\\\\6x=\frac{\pi }{3}-\frac{\pi }{3}+\pi n,n\in z\\\\6x=\pi n,n\in z\\\\x=\frac{\pi n }{6},n\in z$