$\pm$ Careful

$\tan(\cos^{-1}x )=\sin(\tan^{-1}2)$

$\Rightarrow \dfrac{\sqrt{1-x^2}}{x} = \sin(\arctan(2))$

$\Rightarrow \sqrt{1-x^2} = \dfrac{2\sqrt5}{5}x$

$\Rightarrow \left(\sqrt{1 - x^2}\right)^2 = \left(\dfrac{2\sqrt5}{5}x\right)^2$ $~~~~~~$ $[ \text{square on both sides}]$

$\Rightarrow 1 - x^2 = \dfrac{2^2 \times 5}{5^2} x^2$

$\Rightarrow 1 - x^2 = \dfrac45 x^2$

After solving it, we get,

$x = \dfrac{\sqrt{5}}{3}, x = -\dfrac{\sqrt{5}}{3}$

Verifying solutions:

Substituting $x = \dfrac{\sqrt{5}}{3}$ in $\dfrac{\sqrt{1-x^2}}{x}$

$\dfrac{\sqrt{1- \left(\frac{\sqrt 5}{3}\right)^2}}{\frac{\sqrt 5}{3}} = \dfrac{2\sqrt5}{5}$

Substituting $x = -\dfrac{\sqrt{5}}{3}$ in $\dfrac{\sqrt{1-x^2}}{x}$

$\dfrac{\sqrt{1- \left(-\frac{\sqrt 5}{3}\right)^2}}{-\frac{\sqrt 5}{3}} \ne \dfrac{2\sqrt5}{5}$

Hence the final solution is $x = \boxed{\dfrac{\sqrt{5}}{3}}$