<s>[INST] Solve $\tan x =\sqrt{3}$, finding all solutions. My attempt so far:
$$
\tan x = \sqrt3
$$
$$
\frac{\sin x}{\cos x}= \sqrt3 
$$
Then I look at the unit circle to find possible solutions. I find two solutions: $$
\frac{\pi}{3} \text{  and }\frac{4\pi}{3}
$$
However, I find the answer to be:
$$
x = \frac{\pi}{3} + \pi k
$$
Why isn't $4\pi/3$ included in the solution?
(edit: I mistakenly wrote $7\pi/6$ instead of $4\pi/3$. Sorry for the confusion.)
 [/INST] $$\tan\left(\frac {7\pi}{6}\right) = \dfrac{-\frac 12}{-\frac{\sqrt 3}{2}} = \dfrac 1{\sqrt 3}$$
So $x = \dfrac{7\pi}{6}$ does not solve $\tan x = \sqrt 3$.
The second solution $x$ in the interval $[0, 2\pi]$ is $$x = \dfrac{4\pi}3$$
ADDED after EDITED post:
Note that $$\dfrac {4\pi}{3} = \frac{\pi}{3} + 1\cdot \pi = \dfrac{\pi}{3} + k\pi, \;k = 1$$
And we can characterize all solutions $x_i = \pi/3 + k\pi$. 
</s>