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: http://i.stack.imgur.com/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?
Best Answer
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