Answer
$sin~2x=\frac{4\sqrt 5}{9}$
Work Step by Step
$sec~x=\frac{3}{2}$
$cos~x=\frac{1}{sec~x}=\frac{2}{3}$
$sin^2x=1-cos^2x=1-(\frac{2}{3})^2=1-\frac{4}{9}=\frac{5}{9}$
$sin~x=\frac{\sqrt 5}{3}~~$ (Quadrant I)
Use the double-angle formula for sine.
$sin~2x=2~sin~x~cos~x=2~\frac{\sqrt 5}{3}~\frac{2}{3}=\frac{4\sqrt 5}{9}$
