Question

Прошу помогите пожалуйста

Answer 1

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Answer 2

$\displaystyle \sf \boxed{tg x=a;\;\;\;\;\;\;\;\;\; x=arctg \; a+ \pi n, n\in Z}\\\\tg(2x-\frac{\pi}{6})=1\\ 2x-\frac{\pi}{6}=\frac{\pi}{4}+\pi n, n\in Z\\\\ 2x=\frac{\pi}{6}+\frac{\pi}{4}+\pi n, n\in Z\\\\ 2x=\frac{10\pi }{24}+\pi n, n\in Z \;\;\;\; |:2 \\\\ x=\frac{5\pi }{24}+\frac{\pi n }{2}, n\in Z$

Ответ: $\displaystyle \sf \frac{5\pi }{24}+\frac{\pi n }{2}, n\in Z$

$\displaystyle \sf ctg(\frac{x}{2}+\frac{\pi}{8})=\sqrt{3}\\\\ \frac{x}{2}+\frac{\pi}{8}=\frac{\pi}{6}+\pi k, k\in Z \;\;\;\;\;|\cdot 2 \\\\ x=\frac{2\pi}{6}-\frac{2\pi}{8} +2\pi k, k\in Z\\\\ x=\frac{\pi}{12}+2\pi k, k\in Z$

Ответ: $\displaystyle \sf \frac{\pi}{12}+2\pi k, k\in Z$