Answer
\[
\begin{aligned}
x &=\frac{\sqrt{3}}{2} \rho \cos \theta \\
&=\frac{\sqrt{3}}{2} \rho \sin \theta=x \\
&=\frac{1}{2}=z \rho
\end{aligned}
\]
Work Step by Step
In spherical coordinates we have that
\[
\begin{aligned}
&\rho \sin \theta \sin \phi =x\\
&\rho \cos \theta \sin \phi =x\\
&\rho \cos \phi=z
\end{aligned}
\]
Since $\sqrt{x^{2}+y^{2}}=z$
\[
\begin{aligned}
&\sqrt{x^{2}+y^{2}}=z \Rightarrow \rho \cos \phi=\sqrt{3 \rho^{2} \cos ^{2} \theta \sin ^{2} \theta+3 \rho^{2} \sin ^{2} \theta \sin ^{2} \theta} \\
&=\rho \cos \phi=\sqrt{3} \rho \sin \phi \Rightarrow \tan \phi=\sqrt{3} \Rightarrow \phi=\pi / 3
\end{aligned}
\]
So
\[
\begin{aligned}
x &=\frac{\sqrt{3}}{2} \rho \cos \theta \\
&=\frac{\sqrt{3}}{2} \rho \sin \theta=x \\
&=\frac{1}{2}=z \rho
\end{aligned}
\]
