Simplification of formula: $\cos{2\theta} + \sin{2\theta}\tan{\phi}$

Question

So I'm trying to understand a certain proof/derivation of a formula in physics, but I seem to struggle with the following simplification to the final result.

The given formula is:

$$ 0 = \cos{2\theta} + \sin{2\theta}\tan{\phi} $$

And it is simplified to:

$$ \theta = \frac{1}{2}\tan^{-1}{\left(-\frac{1}{\tan{\phi}}\right)} $$

I don't really see how you can get there from the given foruma. I've tried moving things around in the equation but it never gives me the proper result. Anyone have any idea how this is achieved?

Answer 1

Dividing by $\cos(2\theta)$ we obtain

$$ 0 = 1 + \tan(2\theta) \tan \phi $$

Your expression follows

P.S. : We have assumed $\theta \neq \frac{\pi}{4}$, why?

Answer 2

Notice that we can divide by $\cos(2\theta)$ and obtain:

$$-1/\tan\phi=\tan(2\theta)\implies \theta=\tan^{-1}(-1/\tan\phi)/2$$