What am I doing wrongwith trigonometric manipulation in this question?

Question

I am working on the following problem:

Evaluate the domain of: $$y = \frac{x}{\sqrt{\sin({lnx}) - \cos({lnx})}} $$

Here is how I proceeded : For $\ln{x}$ to be define, $x>0$ (Condition 1)

Secondly, $\sin({lnx}) - \cos({lnx}) > 0 $

It implies :

$=> \sin({lnx})>\cos({lnx}) $

$=> \tan({lnx})>1 $ (STEP 3)

$=>\pi/4 +\pi n < ln x < \pi/2 + \pi n$ wherer $\pi n$ = period of $\tan{x}$

$=>e^{\pi/4 +\pi n} < x < e^{\pi/2 +\pi n}$

But the answer of the book is quite different from mine and it is correct(no typo),I verified it by graphing the function on Desmos.

BOOK ANSWER: $=>e^\pi(1/4 + 2n) < x < e^\pi(5\pi/4 + 2n)$

(^Apologies for improper rendering. Kindly, take bracketed expression as exponent along with $\pi$

And here is how the book attempts it :

$\sin({lnx}) - \cos({lnx}) > 0 $

$=>\sin({lnx}) > \cos({lnx})$

$=>\pi/4 + 2\pi n < ln x < 5\pi/4 + 2\pi n $

Therefore, $=>e^\pi(1/4 + 2n) < x < e^\pi(5\pi/4 + 2n)$ (ANSWER)

I want to know what am I doing worng here, am I overlooking some importnat fact while carrying out some steps/operations. in STEP 3

Answer 1

Your Step 3 is not valid if $\cos({ln{x}})<0$. If you divide an inequality by a negative number the sign changes.