Giải các phương trình sau:

a) tan x = 1;

b) tan x = –1;

c) tan x = 0.

a) Ta có tan x = 1

$\iff \text{tanx}=\tan \frac{\pi }{4}$

$\iff \text{x}=\frac{\pi }{4}+k\pi ,k\in \mathbb{Z}$

Vậy $\iff \text{x}=\frac{\pi }{4}+k\pi ,k\in \mathbb{Z}$ .

b) Ta có tan x = –1

$\iff \text{tanx}=\tan \left(-\frac{\pi }{4}\right)$

$\iff \text{x}=-\frac{\pi }{4}+k\pi ,k\in \mathbb{Z}$

Vậy $\iff \text{x}=-\frac{\pi }{4}+k\pi ,k\in \mathbb{Z}$ .

c) tan x = 0

⇔ tanx = tan0

⇔ x = kπ, k ∈ Z

Vậy x = kπ, k ∈ Z.