Write the equation $4x^{2}+4z^{2}=5$ in spherical coordinates. Write the equation $4x^{2}+4z^{2}=5$ in spherical coordinates.
I used the facts that
$$
\begin{align}
x&=ρ\sin\theta\cos\phi\;,\\
z&=ρ\cos\phi\;,
\end{align}
$$
And ended up with:
$ 4 (ρ^2 \sin^2(\phi) \cos(\theta) + ρ^2\cos^2(\phi))= 5 $
But it's not simplified enough? I can't use Pythagoras' theorem: $\cos^2(\theta) + \sin^2(\theta) = 1$ inside the parenthesis.
So what do?
 A: It depends how you define spherical coordinates that ultimately determines what the equation looks like. If you define spherical coordinates like this 
$$x = \rho \sin \theta \cos \phi$$
$$y = \rho \sin \theta \sin \phi$$
$$z = \rho \cos \theta$$
where $\rho$ is the radius, $\theta$ is the inclination, and $\phi$ is the azimuth. Naturally, $\rho \in [0, \infty), \> \theta \in [0, \pi], \> \phi \in [0, 2 \pi]$. 
(notice that the cosines are different for x and z unlike in your definition, although I think this is what you meant) 
then 
$$\rho^2(\cos^2 \phi \sin^2 \theta + \cos^2 \theta) = \frac{5}{4} $$
$$\rho^2(\cos^2 \phi \sin^2 \theta + 1 - \sin^2 \theta) = \frac{5}{4} $$
$$\rho^2(\sin^2\theta(\cos^2 \phi -1) + 1) = \frac{5}{4}$$
$$\rho^2(1 - \sin^2 \theta \sin^2 \phi) = \frac{5}{4}$$
is the simplest. However, you do not have to define spherical coordinates in such a manner.
This convention is used more often in physics where in mathematics you would often see $\theta$ and $\phi$ be flipped in definition.
