Answer
$-\dfrac{5}{13}$
Work Step by Step
$\alpha = \tan^{-1} {\dfrac{3}{2}} $
$\tan{\alpha} = \dfrac{3}{2}$
$\sec{\alpha} =\sqrt{1+\tan^2{\alpha}} = \dfrac{\sqrt{13}}{2}$
$\cos{\alpha} = \dfrac{1}{\sec{\alpha}} = \dfrac{2}{\sqrt{13}}$
$\cos{2\alpha} =2\cos^2{\alpha} -1 = -\dfrac{5}{13}$