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**Hint:**Domain of a real valued function is a set of real values of x for which the function y is defined. Assume that the given function is a real valued function (from a real set to a real set of numbers). Then analyse the given function and find out for which value of x, y is a real number.

**Formula used:**

$\cos x\sin y-\sin x\cos y=\sin \left( x-y \right)$

**Complete step by step solution:**

Let us first understand what is the domain of a function. Domain of a real valued function is a set of real values of x for which the function y is defined. Here, the function is given to be

$y=\dfrac{x}{\sqrt{\sin (\ln x)-\cos (\ln x)}}$

Now, we have to find those real values of x for which the value of y exists.We can see that the numerator of the function is x, which is always real. However, the denominator contains a square root. The number under a square root must be always a positive real number. Otherwise the result will be an imaginary number which is not real. This means that for the real value of y, $\sin (\ln x)-\cos (\ln x)>0$ .

Now, multiply and divide the above inequality by $\sqrt{2}$ as shown below.

$\sqrt{2}\left( \dfrac{1}{\sqrt{2}}\sin (\ln x)-\dfrac{1}{\sqrt{2}}\cos (\ln x) \right)>0$

We know that $\dfrac{1}{\sqrt{2}}=\cos \left( \dfrac{\pi }{4} \right)=\sin \left( \dfrac{\pi }{4} \right)$.

Then, we can write the inequality as

$\sqrt{2}\left( \cos \dfrac{\pi }{4}\sin (\ln x)-\sin \dfrac{\pi }{4}\cos (\ln x) \right)>0$

Let us consider that $\sqrt{2}$ > 0

Then,

$\Rightarrow \cos \dfrac{\pi }{4}\sin (\ln x)-\sin \dfrac{\pi }{4}\cos (\ln x)>0$

Now, we can write that $\cos \dfrac{\pi }{4}\sin (\ln x)-\sin \dfrac{\pi }{4}\cos (\ln x)=\sin \left( \ln x-\dfrac{\pi }{4} \right)$.

With this we get that $\sin \left( \ln x-\dfrac{\pi }{4} \right)>0$

We know that sine function is positive when the angle is in the first and second quadrant.The general notation for the angles that fall in the first and second quadrant is $n\pi <\theta <(n+1)\pi $, where $\theta $ is the angle and n belongs to integers.

Then this means that $n\pi <\ln x-\dfrac{\pi }{4}<(n+1)\pi $

$\Rightarrow n\pi +\dfrac{\pi }{4}<\ln x<(n+1)\pi +\dfrac{\pi }{4}$

$\Rightarrow \dfrac{(4n+1)\pi }{4}<\ln x<\dfrac{(4n+5)\pi }{4}$

$\therefore {{e}^{\dfrac{(4n+1)\pi }{4}}} < x< {{e}^{\dfrac{(4n+5)\pi }{4}}}$.

**Therefore, the domain of the given function is ${{e}^{\dfrac{(4n+1)\pi }{4}}} < x < {{e}^{\dfrac{(4n+5)\pi }{4}}}$.**

**Note:**Some students may argue that the number under square root can be zero as square root of zero is zero and it is a real number. However, observe that the square root is in the denominator and for a real value of y to exist, the denominator cannot be equal to zero.

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