## Algebra and Trigonometry 10th Edition

$x=\frac{\pi}{3}+n\pi$ and $x=\frac{2\pi}{3}+n\pi$, where $n$ is an integer.
$3~cot^2x-1=0$ $3~cot^2x=1$ $cot^2x=\frac{1}{3}$ $cot~x=±\frac{1}{\sqrt 3}=±\frac{1}{\sqrt 3}\frac{\sqrt 3}{\sqrt 3}=±\frac{\sqrt 3}{3}$ The period of $cot~x$ is $\pi$. The solutions in the interval: $[0,\pi)$ is: $x=\frac{\pi}{3}$ and $x=\frac{2\pi}{3}$ Now, add multiples of $\pi$ to the solution: $x=\frac{\pi}{3}+n\pi$ and $x=\frac{2\pi}{3}+n\pi$, where $n$ is an integer.