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.)
$\endgroup$ 43 Answers
$\begingroup$$tan(x) = \sqrt 3$ is positive in the first and third quadrants. The principal value for this equation is $\frac {\pi}{3}$; the second value in the third quadrant is $\frac {4 \pi}{3}$ (valid in $0<x<2 \pi$); the general value is $\frac {\pi}{3} + \pi k$.
$\endgroup$ $\begingroup$$$\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 $x_i = \pi/3 + k\pi$.
$\endgroup$ 2 $\begingroup$In the answer, if k=1 you get the solution $ \frac{{4\pi }}{3} $
$\endgroup$ 3
