The lateral surface area of a right circular cone is three times the surface area of its inscribed sphere. Find the vertex angle (cone)

I am quite confused on how to attack this problem... It's not homework I tried to use the equations of the surface area of a sphere and the lateral area of the cone to find the answer, but it didn't work... The image is just showing how a sphere is inscribed inside the cone. Thanks! *by the vertex angle I meant the pheta shown in the image Draw a line from the center of the sphere to the lower rim, and let $\alpha$ be the angle it makes with the diameter ending there. Then $a=r\tan\alpha$, and $\theta={\pi\over2}-2\alpha$.
Denote the distance from the vertex to the rim by $s$. Then $$s={r\over\sin\theta}={r\over\cos(2\alpha)}\ .$$
The area of the lateral surface of the cone is given by $$A_{\rm lat}={1\over2}\>2\pi r\>s={\pi r^2\over\cos(2\alpha)}\ ,$$ and area of the sphere is given by $$A_{\rm sph}=4\pi a^2=4\pi r^2\tan^2\alpha\ .$$ The condition $A_{\rm lat}=3A_{\rm sph}$ then leads to the equation $$12\cos(2\alpha)\tan^2\alpha=1\ .$$ Letting $$\sin\theta=\cos(2\alpha)=:u$$ we have $\tan^2\alpha={1-u\over1+u}$, so that we now have to solve $$12u^2-11u+1=0\ .$$ There are two solutions $u_{\pm}={1\over24}(11\pm\sqrt{73})$, both leading to an acceptable $\theta$-value $\theta_\pm=\arcsin u_{\pm}$.