Find a root of f(x) = 0, arccos & arcsin

Question

Can someone please help me with this question?

Let $f(x) = 2\arccos(\frac{x}{2}) + 6\arcsin(\frac{3}{2x}) - 2 \pi$

Find a root of $f(x) = 0$, that is a point x where $f(x) = 0$.

Answer 1

Here is what I have so far,

Let $ \frac{x}{2} = \cos \theta $ and $ \frac{3}{2x} = \sin \phi $, now we have a much simpler equation to work on.

$ 2\theta + 6\phi - 2\pi = 0 $

Just some rearrangement we get $ \theta = \pi - 3\phi $.

Next, let eliminate $ \theta $, now we have

$ 2 \cos(\pi - 3\phi) = 2 \cos \theta = x = \frac{3}{2 \sin \phi} $

$ \cos(\pi - 3\phi)\sin \phi = \frac{3}{4} $

$ \cos(3\phi)\sin \phi = -\frac{3}{4} $

Now we apply the product to sum formula and get

$ \frac{1}{2}(\sin 4\phi - \sin 2\phi) = -\frac{3}{4} $

$ (\sin 4\phi - \sin 2\phi) = -\frac{3}{2} $

Solving the equation above numerically, we get

$ \phi \approx 0.928 $

So $ x = \frac{3}{2 \sin \phi} \approx 1.87 $

That's a root of the original equation!