Answer
$(2,\frac{\pi}{3},\frac{\pi}{6})$.
Work Step by Step
Since $ x=\frac{1}{2}, \, y=\frac{\sqrt3}{2}, \, z=\sqrt 3$ and
$$\rho=\sqrt{x^2+y^2+z^2},\quad \theta=\tan^{-1}( y/x), \quad \phi=\cos^{-1}(z/\rho),$$
then we have
$$\rho=\sqrt{\frac{1}{4}+\frac{3}{4}+3}= 2, $$
$$\theta=\tan^{-1}( \sqrt 3)=\frac{\pi}{3}, $$
$$\phi=\cos^{-1}(\sqrt 3/2)=\frac{\pi}{6}.$$
Hence, the spherical coordinates are $(2,\frac{\pi}{3},\frac{\pi}{6})$.
