## Precalculus (10th Edition)

$2.08,5.22$.
$5\tan{\theta}+9=0\\5\tan{\theta}=-9\\\tan{\theta}=-\frac{9}{5}$ The domain of $\tan^{-1}{x}$ is the range of $\tan{x}$, which is the real numbers. The range of $\tan^{-1}{x}$ is a specified domain of $\tan{x}$, i.e. $\left(− \frac{\pi}{2} , \frac{\pi}{2} \right)$. This equation is the same as: $tan(\theta)=-\frac{9}{5}.$ After using a calculator in radian mode the solutions for $\tan^{-1}{-\frac{9}{5}}$ is $-1.06$ (given by the calculator) and because the tan function has a period of $\pi$, the first solution is $-1.06+\pi=2.08$ and the second solution is $-1.06+2\pi=5.22.$