Answer
$sin2x=\frac{12}{13}$
$cos2x=-\frac{5}{13}$
$tan2x=-\frac{12}{5}$
Work Step by Step
$tan(x)=\frac{3}{2}, sin(x)=\frac{3\sqrt {13}}{13}, cos(x)=\frac{2\sqrt {13}}{13}$
$sin2x=2sin(x)cos(x)=\frac{12}{13}$
$cos2x=2cos^2x-1=-\frac{5}{13}$
$tan2x=\frac{sin2x}{cos2x}=-\frac{12}{5}$