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lukyo

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Find the domain of the function:

\(\mathsf{f(x)=\dfrac{cos\,x}{1-sin\,x}}\)


Denominators can't be equal to zero:

\(\mathsf{1-sin\,x e 0}\\\\ \mathsf{sin\,x e 1}\\\\ \mathsf{sin\,x e sin\,\dfrac{\pi}{2}}\)


So,

\(\mathsf{x e \dfrac{\pi}{2}+k\cdot 2\pi,\qquad\quad k\in\mathbb{Z}.}\)


Domain:

\(\mathsf{Dom(f)=\left\{x\in\mathbb{R}:~x e \frac{\pi}{2}+k\cdot 2\pi,~~k\in\mathbb{Z}\right\}.}\)


I hope this helps. =)

The function's domain is indeed the set among all features for the function. The domain would be all the values in a method and all the outgoing values are within range. Specific functions with much more limited domains can also be described.  

Following are calculations of the domain:

Given:

\(\bold{\to f(x)= \frac{(\cos x)}{(1-\sin x)}}\)

Find:

Domain=?

Solution:

The functions and periods for \(\bold{ \frac{\cos \left(x\right)}{1-\sin \left(x\right)}}\)  are as follows:

\(\cos x\) with \(2 \pi\) periodicity.

\(\sin x\)  with \(2 \pi\) periodicity.

The compound periodicity is therefore \(2 \pi\):

\(Domain \ of =\frac{\cos x}{1-sin x} \begin{bmatrix}Solution & 2\pi n\le \:x<\frac{\pi }{2}+2\pi n \ Or \ \frac{\pi }{2}+2\pi n

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