So here's the deal. I'll put the logarithm bases into (). Solve log(cosx)sinx+ log(sinx)cosx=2.

Question

anonymous · Answer

Whew, the bases are functions of x? That's pretty gnarly.
Mixing logs and trig functions . . . will probably end up with complex numbers.
This will take some work.

anonymous · Answer

It should be done without complex numbers.

anonymous · Answer

My best guess is pi/4.

anonymous · Answer

There may be other solutions, but I know that works.

anonymous · Answer

Want to know how I did it?

anonymous · Answer

Sure.

anonymous · Answer

I took the simplest case in which 
$$\log_{\cos x}(\sin x) =\log_{\sin x}(\cos x) = 1.$$
A logarithm equals 1 when the base and what it is operating on are equal.
sin x = cos x at pi/4.

anonymous · Answer

That's true.

anonymous · Answer

It'll take a little more doing to see if there are other solutions.

anonymous · Answer

e.g. log_{cos x}(sin x) could be 1.5 and log_{sin x}(cos x) could be 0.5 and so on.

anonymous · Answer

That's also true.

anonymous · Answer

I think I can continue from now on. You gave me the ideas I needed. Thank you very much!

anonymous · Answer

Thank you. That was a fun one!