Answer
$x = \frac{1}{3}~(tan^{-1}~2y - 2)$
where $y$ is in the interval $(-\infty, \infty)$
Work Step by Step
$y = \frac{1}{2}~tan~(3x+2)$
$x$ is restricted to the interval $(-\frac{2}{3}-\frac{\pi}{6}, -\frac{2}{3}+\frac{\pi}{6})$
We can find $y$ when $x$ approaches the value $~~-\frac{2}{3}-\frac{\pi}{6}~~$ from the right:
$y = \frac{1}{2}~tan~(3x+2)$
$y = \frac{1}{2}~tan~[3( -\frac{2}{3}-\frac{\pi}{6})+2]$
$y = \frac{1}{2}~tan~[-2-\frac{\pi}{2}+2]$
$y = \frac{1}{2}~tan~(-\frac{\pi}{2})$
$y = -\infty$
We can find $y$ when $x$ approaches the value $~~-\frac{2}{3}+\frac{\pi}{6}~~$ from the left:
$y = \frac{1}{2}~tan~(3x+2)$
$y = \frac{1}{2}~tan~[3( -\frac{2}{3}+\frac{\pi}{6})+2]$
$y = \frac{1}{2}~tan~[-2+\frac{\pi}{2}+2]$
$y = \frac{1}{2}~tan~(\frac{\pi}{2})$
$y = \infty$
When x is restricted to the interval $(-\frac{2}{3}-\frac{\pi}{6}, -\frac{2}{3}+\frac{\pi}{6})$, $y$ has a value on the interval $(-\infty, \infty)$
We can find an expression for $x$:
$y = \frac{1}{2}~tan~(3x+2)$
$tan^{-1}~2y = 3x+2$
$x = \frac{1}{3}~(tan^{-1}~2y - 2)$
where $y$ is in the interval $(-\infty, \infty)$
