Trigonometric problem using basic trigonometry 
If $x$ is a solution of the equation: $$\tan^3 x = \cos^2 x - \sin^2 x$$
  Then what is the value of $\tan^2 x$?

This is the problem you are supposed to do it just with highschool trigonometry , but i can't manage to do it please help
Here are the possible answers:
$$a) \sqrt{2}-1, b) \sqrt{2}+1, c) \sqrt{3}-1, d) \sqrt{3}+1, e)\sqrt{2}+3$$

$\tan^2 x=\cos^2 x-\sin^2 x$
$\sin^2 x=\cos^4 x -\sin^2 x \cos^2 x$
$0=\cos^4 x -\sin^2 x \cos^2 x -\sin^2 x$
$0=\cos^4 x - \sin^2 x(1+\cos^2 x)$
$0=\cos^4 x - (1-\cos^2 x)(1+\cos^2 x)$
$0=\cos^4 x -(1- \cos^4 x)$
$0=2 \cos^4 x -1$
$\cos^2 x=\frac {\sqrt 2}{2}$
$\large \frac {1}{\cos^2 x}=\sqrt 2$
$\tan^2 x=\sqrt 2 \sin^2 x$
Looks like we're stuck, but from above we have $\cos^2 x=\frac {\sqrt 2}{2}$ , so $\sin^2 x=1-\frac {\sqrt 2}{2}$
By substitution, $\tan^2 x=\sqrt 2 (1-\frac {\sqrt 2}{2})$
$\tan^2 x=\sqrt 2 -1$
