Precalculus (6th Edition) Blitzer

The solutions of the equation are $x=\frac{\pi }{3}+n\pi$, where n is any integer.
The x in $\tan x=\sqrt{3}$. Because $\tan \frac{\pi }{3}=\sqrt{3}$, and the solution of $\tan x=\sqrt{3}$ in [0, $\pi$ ) is: $x=\frac{\pi }{3}$ The period of the tangent function is $\pi$, and the solutions are given by: $x=\frac{\pi }{3}+n\pi$, where n is any integer.