How to solve this equation in spherical coordinates

Question

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.

Answer 1

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.

Answer 2

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 $z=\sqrt{R^2-r^2}\pm 1$.

Answer 3

(BIG) HINT:

$$\cos(\phi)+\sqrt{\cos^2(\phi)+15}=\frac{2}{\sin(\phi)}\Longleftrightarrow$$ $$\cos(\phi)+\sqrt{\cos^2(\phi)+15}=2\csc(\phi)\Longleftrightarrow$$ $$\sqrt{\cos^2(\phi)+15}=2\csc(\phi)-\cos(\phi)\Longleftrightarrow$$ $$\cos^2(\phi)+15=\left(2\csc(\phi)-\cos(\phi)\right)^2\Longleftrightarrow$$ $$\cos^2(\phi)+15=\cos^2(\phi)-4\cot(\phi)+4\csc^2(\phi)\Longleftrightarrow$$ $$15+4\cot(\phi)-4\csc^2(\phi)=0\Longleftrightarrow$$ $$11+4\cot(\phi)-4\cot^2(\phi)=0\Longleftrightarrow$$ $$-\frac{11}{4}-\cot(\phi)+\cot^2(\phi)=0\Longleftrightarrow$$ $$\cot^2(\phi)-\cot(\phi)=\frac{11}{4}\Longleftrightarrow$$ $$\left(\cot(\phi)-\frac{1}{2}\right)^2=3\Longleftrightarrow$$ $$\cot(\phi)-\frac{1}{2}=\pm\sqrt{3}\Longleftrightarrow$$ $$\cot(\phi)=\pm\sqrt{3}+\frac{1}{2}\Longleftrightarrow$$ $$\cot(\phi)=\frac{1}{2}\pm\sqrt{3}$$