## Precalculus (6th Edition) Blitzer

The solution of $\tan x=-\sqrt{3}$ in $[0,\pi )$ is $x=\pi -\frac{\pi }{3}$, or $x=\underline{\frac{2\pi }{3}}$. If n is any integer, all solutions of $\tan x=-\sqrt{3}$ are given by $\frac{2\pi }{3}+n\pi$.
To calculate another value of x: \begin{align} & x=\pi -\frac{\pi }{3} \\ & =\frac{3\pi -\pi }{3} \\ & =\frac{2\pi }{3} \end{align} Therefore, the value of x will be $\frac{2\pi }{3}$. The interval $[0,\pi )$ covers the 1st and 2nd quadrants. The value of x is $\frac{2\pi }{3}$, according to the standard trigonometric general solution. Since, all the solutions for $\tan x=-\sqrt{3}$ can be written as $\frac{2\pi }{3}+n\pi$. The value is considered to be $\pi$ because the interval has mentioned the limit of $[0,\pi )$. Thus, the values of x will be $\frac{2\pi }{3}\,\text{ and }\,\frac{2\pi }{3}+n\pi$.