Find the equation in spherical coordinates of $x^2 + y^2 – z^2 = 4$. Find the equation in spherical coordinates of $x^2 + y^2 – z^2 = 4$.
$$\begin{align}
x^2 + y^2 &= r^2\sin^2(\theta)\\
z^2 &= r^2 \cos(\theta) \\
x^2 + y^2 - z^2&=r^2(\sin^2(\theta) - \cos^2(\theta)) = 4
\end{align}$$
Thus, $$-r^2(\cos(2\theta)) = 4$$
Is this right?

The equation given is correct, but we can simplify further.  Multiply by $-\sec 2\theta$ and take the square root to solve for $r$ which is $>0$.  Thus $r=2\sqrt{-\sec(2\theta)}$.
Note that the result appears to be the square root of a negative number ... unless the secant function is negative.  Recalling where the cosine function and thus the secant function is negative, conclude that $\pi/4<\theta<3\pi/4$.  This goes along with the hyperboloid shape you might recognize from the rectangular equation.
