# The domain of logarithmic functions with sinx and cosx

I need to solve this equation:

$\log_{\cos x} \sin x + \log_{\sin x} \cos x=2$

But in order to solve it, I first need to find the domain. What I did was this:

$\cos x\neq1 \wedge \sin x\neq1$ which results in $x\neq2k\pi$ and $x\neq\frac{\pi}{2}+2k\pi$

$\cos x>0$ therefore $x\in(\frac{\pi}{2}+2k\pi, \frac{3\pi}{2}+2k\pi)$

$\sin x>0$ therefore $x\in(2k\pi, \pi+2k\pi)$

Correct me if I made a mistake here. Anyway, how do I make the domain out of these conditions?

The solution says the domain is this: https://i.sstatic.net/FToOP.png

The domain affects the solution $x\in\frac{\pi}{4}+k\pi$ by turning it into $x\in\frac{\pi}{4}+2k\pi$.

So, my question is, how do I join those conditions (if they're complete and correct), and how do I determine which part of the solution fits the domain?

• It might be a little easier to think first about finding solutions in the first period $(0,2\pi)$ and then adding any integer multiple of $2\pi$ to those solutions, rather than carry the periodic framework through the solution process from beginning to end. Commented May 27, 2015 at 16:55

Clearly we need $0<\sin x,\cos x<1$
All Sin Tan Cos Rule says: we need $x$ to be in the first Quadrant i.e., $2m\pi<x<2m\pi+\dfrac\pi2$ where $m$ is any integer
• @Quant, In the real finite algebra, the base can be $(0,1)\cup(1,\infty)$ Commented May 27, 2015 at 17:00