Задание 16.11

Question

$^{\circ}$ Решите уравнение:
$$1) \operatorname{tg}\left(3 x-\frac{\pi}{12}\right)=\frac{\sqrt{3}}{3}$$
$$2) \operatorname{tg}(3-2 x)=2$$

Gimli · Accepted Answer

$$\operatorname{tg}\left(3 x-\frac{\pi}{12}\right)=\frac{\sqrt{3}}{3}$$
$$3 x-\frac{\pi}{12}=\operatorname{arctg} \frac{\sqrt{3}}{3}+\pi n$$
$$3 x-\frac{\pi}{12}=\frac{\pi}{3}+\pi n$$
$$3 x=\frac{\pi}{3}+\frac{\pi}{12}+\pi n$$
$$x=\frac{\pi}{9}+\frac{\pi}{36}+\frac{\pi n}{3}, n \in 2$$
2) $$\operatorname{tg}(3-2 x)=2$$
$$3-2 x=\operatorname{arctg} 2+\pi n$$
$$-2 x=\operatorname{arctg} 2+\pi n-3$$
$$x=-\frac{\operatorname{arctg} 2}{2}+\frac{\pi n}{2}+1,5$$

Задание 16.11

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Задание 16.11

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