Answer: A
\[ \sin x = \frac{2}{3} \] \[ \therefore \sin^2 x = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \] So, \[ 1 - \sin^2 x = 1 - \frac{4}{9} = \frac{5}{9} \] That means, \[ \cos^2 x = \frac{5}{9} \] So, \[ \cos x = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \] \[ \therefore \tan x = \frac{\sin x}{\cos x} = \frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}} = \frac{2}{\sqrt{5}} \]
\[ \sin x = \frac{2}{3} \] \[ \therefore \sin^2 x = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \] So, \[ 1 - \sin^2 x = 1 - \frac{4}{9} = \frac{5}{9} \] That means, \[ \cos^2 x = \frac{5}{9} \] So, \[ \cos x = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \] \[ \therefore \tan x = \frac{\sin x}{\cos x} = \frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}} = \frac{2}{\sqrt{5}} \]