Задание 16.12

Question

$^{\circ}$ Решите уравнение:
$$1) \operatorname{tg}\left(x+\frac{\pi}{4}\right)=1$$
$$2) 3 \operatorname{tg}(3 x+1)-\sqrt{3}=0$$.

Gimli · Accepted Answer

Решение:
ewline
1)$$\operatorname{tg}\left(x+\frac{\pi}{4}ight)=1 $$
$$ x+\frac{\pi}{4}=\operatorname{arctg} 1+\pi n$$
 $$x+\frac{\pi}{4}=\frac{\pi}{4}+\pi n$$
$$  x=\pi n, 	ext{ где }n\in \mathbb{Z}$$
2) $$3 \operatorname{tg}(3 x+1)-\sqrt{3}=0 $$
$$ \operatorname{tg}(3 x+1)=\frac{1}{\sqrt{3}}$$
$$3 x+1=\operatorname{arctg} \frac{\sqrt{3}}{3}+\pi n$$
$$ 3 x+1=\frac{\pi}{6}+\pi n$$
$$x=\frac{\pi}{18}+\frac{\pi n}{3}-\frac{1}{3}, n \in \mathbb{Z}$$

Задание 16.12

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Задание 16.12

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