Draw a right triangle with legs $1, x$; then the angle $\theta$ opposite $x$ is $\tan^{-1} x$, and we can compute $\cos (\theta)=1 / \sqrt{x^{2}+1}$. Thus, we only need to solve $x=1 / \sqrt{x^{2}+1}$. This is equivalent to $x \sqrt{x^{2}+1}=1$. Square both sides to get $x^{4}+x^{2}=1 \Rightarrow x^{4}+x^{2}-1=0$. Use the quadratic formula to get the solution $x^{2}=(-1+\sqrt{5}) / 2$ (unique since $x^{2}$ must be positive).