Answer
$y=2x-\frac{\pi}{2}+1$ and $y=2x+\frac{\pi}{2}-1$.
See graph.
Work Step by Step
Step 1. Given $y=tan(x)$, we have $y'=sec^2x$
Step 2. For a tangent line parallel to the line $y=2x$, let $y'=2$, we have $sec^2x=2$ and $sec(x)=\pm\sqrt 2$ thus $x=\pm\pi/4$ in $(-\pi/2,\pi/2)$
Step 3. At $x=\pi/4$, $y=tan(\pi/4)=1$, and the tangent line equation is $y-1=2(x-\pi/4)$ or $y=2x-\frac{\pi}{2}+1$
Step 4. At $x=-\pi/4$, $y=tan(-\pi/4)=-1$, and the tangent line equation is $y+1=2(x+\pi/4)$ or $y=2x+\frac{\pi}{2}-1$
Step 5. See graph.
