0
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – hardmath
    Commented May 27, 2015 at 16:55

1 Answer 1

Reset to default
1
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ cosx should not be less than one, because it is in the logarithm. Can you clarify your solution, I didn't really understand it? $\endgroup$
    – Quant
    Commented May 27, 2015 at 16:59
  • $\begingroup$ @Quant, In the real finite algebra, the base can be $(0,1)\cup(1,\infty)$ $\endgroup$
    – lab bhattacharjee
    Commented May 27, 2015 at 17:00
  • $\begingroup$ That's right, thank you! $\endgroup$
    – Quant
    Commented May 27, 2015 at 17:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .