Question

решить (без константы и прочего лишнего)

Answer 1

$tg(5x+\frac{\pi }{4})=-1\\\\5x+\frac{\pi }{4}=arc tg(-1)+\pi n,n\in Z\\\\5x+\frac{\pi }{4}=-\frac{\pi }{4}+\pi n,n\in Z\\\\5x=-\frac{\pi }{2}+\pi n,n\in Z\\\\\boxed{x=-\frac{\pi }{10}+\frac{\pi n }{5},n\in Z }$