I have the following inequation: $$\log_{\cos(x)} \sin(x) + \log_{\sin(x)} \cos(x) \le 2$$ I know that $\sin(x)$ and $\cos(x)$ will give values in the interval $[-1, 1 ]$ but in the base there can't be any numbers $x \lt 0$ and in the argument can only be $\Bbb R^+ \backslash \{ 0 \}$. This being inequation makes it even harder and still after a few hours I'm clueless
How can you solve this and how would you calculate this function domain?
Note: The inequation/function is indeed written down correctly with $\le$ and it's possible there are no solutions.
Note that
$$\log_ab=\frac{\ln b}{\ln a}$$
By, AM-GM, we get
$$\log_{\cos(x)} \sin(x) + \log_{\sin(x)} \cos(x) =\frac{\ln \sin x}{\ln \cos x}+\frac{\ln \cos x}{\ln\sin x}\ge 2$$
If the OP is written down correctly, then the inequality becomes equality, namely $$\log_{\cos(x)} \sin(x) + \log_{\sin(x)} \cos(x) =\frac{\ln \sin x}{\ln \cos x}+\frac{\ln \cos x}{\ln\sin x}= 2$$
therefore,
$$\ln\sin x=\ln\cos x\Longrightarrow \ln\tan x=0\Longrightarrow \tan x=1$$
so the solution is
$$x=\frac\pi4+n\pi,~~~n\in\mathbb Z$$