How to solve this equation in spherical coordinates I am trying to find the angles $\phi$ that satisfy the following equation:
$$
\cos\phi + \sqrt{\cos^2\phi+15}=\frac{2}{\sin\phi},
$$
where $\phi \in [0,\pi ]$.
The geometric interpretation of this equation is the intersection between the sphere $x^2+y^2+(z-1)^2=16$ and the cylinder $x^2+y^2=4$.
To solve this, I have shifted the origin $1$ unit above $(0,0,0)$, in order for the sphere to be centered at the origin. Since the cylinder is infinitely long, the equation is than trivial, giving in this new basis angles $\pi/6$ and $5\pi/6$.
How can I solve the initial equation, or eventually find its solution from angles $\pi/6$ and $5\pi/6$?
Any other approach is welcomed. 
 A: (BIG) HINT:
$$\cos(\phi)+\sqrt{\cos^2(\phi)+15}=\frac{2}{\sin(\phi)}\Longleftrightarrow$$
$$\cos(\phi)+\sqrt{\cos^2(\phi)+15}=2\csc(\phi)\Longleftrightarrow$$
$$\sqrt{\cos^2(\phi)+15}=2\csc(\phi)-\cos(\phi)\Longleftrightarrow$$
$$\cos^2(\phi)+15=\left(2\csc(\phi)-\cos(\phi)\right)^2\Longleftrightarrow$$
$$\cos^2(\phi)+15=\cos^2(\phi)-4\cot(\phi)+4\csc^2(\phi)\Longleftrightarrow$$
$$15+4\cot(\phi)-4\csc^2(\phi)=0\Longleftrightarrow$$
$$11+4\cot(\phi)-4\cot^2(\phi)=0\Longleftrightarrow$$
$$-\frac{11}{4}-\cot(\phi)+\cot^2(\phi)=0\Longleftrightarrow$$
$$\cot^2(\phi)-\cot(\phi)=\frac{11}{4}\Longleftrightarrow$$
$$\left(\cot(\phi)-\frac{1}{2}\right)^2=3\Longleftrightarrow$$
$$\cot(\phi)-\frac{1}{2}=\pm\sqrt{3}\Longleftrightarrow$$
$$\cot(\phi)=\pm\sqrt{3}+\frac{1}{2}\Longleftrightarrow$$
$$\cot(\phi)=\frac{1}{2}\pm\sqrt{3}$$
A: As the whole situation has a rotational symmetry w.r.t. to $z$ we can simplify the problem by looking at a plane which goes throu the z-axis.
In this image you can see that $R=4$ is the radius of the sphere, $r=2$ is the radius of the cylinder. You can also see that the intersectins are circles. And In this case you can easily find that those circles have radius $2$, and are (via pythagoras) at $z=\sqrt{R^2-r^2}\pm 1$.

A: I would write it as
$$
\sqrt{\cos^2\phi+15} = \frac{2}{\sin \phi} - \cos\phi
$$
Both left and right hand side are positive (where defined), so you can square both sides. You will get to
$$
\cos^2\phi + 15 = \frac{4}{\sin^2\phi} + \cos^2\phi - 4 \frac{\cos\phi}{\sin\phi}.$$
Cancel what you can, multiply by $\sin^2\phi$, and use double angle formulas to get everything in terms of $\sin 2\phi$ and $\cos 2\phi$.
If you want to see more steps done, read further.
After canceling and multiplying by $\sin^2\phi$ you get
$$
15\sin^2\phi + 4\cos\phi\sin\phi - 4 = 0
$$
Using double-angle formulas, you get
$$
15\frac{1-\cos 2\phi}{2} + 2\sin 2\phi -4 = 0.
$$
Hence
$$
15\cos 2\phi + 4\sin 2\phi +7 = 0.
$$
From here, you have to solve a linear equation in the unknown $2\phi$. This kind of equation can be solved following a recipe, as explained here.
