## Algebra and Trigonometry 10th Edition

$x=n\pi$ $x=arctan~2+n\pi$ where $n$ is an integer.
$tan^2x-2~tan~x=0$ $tan~x(tan~x-2)=0$ $tan~x=0$ or $tan~x=2$ $tan~x=0$ $x=0$ $tan~x=2$ $x=arctan~2$ The period of $tan~x$ is $\pi$. The general solutions are: $x=0+n\pi=n\pi$ $x=arctan~2+n\pi$ where $n$ is an integer.