Question

find the domain of f(x)=(cos x)/(1-sinx)

Answer 1

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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:

$\mathsf{1-sin\,x\ne 0}\\\\ \mathsf{sin\,x \ne 1}\\\\ \mathsf{sin\,x \ne sin\,\dfrac{\pi}{2}}$


So,

$\mathsf{x\ne \dfrac{\pi}{2}+k\cdot 2\pi,\qquad\quad k\in\mathbb{Z}.}$


Domain:

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


I hope this helps. =)

Answer 2

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 $2 \pi$:

$Domain \ of =\frac{\cos x}{1-sin x} \begin{bmatrix}Solution & 2\pi n\le \:x<\frac{\pi }{2}+2\pi n \ Or \ \frac{\pi }{2}+2\pi n<x<2\pi +2\pi n\\ Interval \ Notation & (2\pi n,\:\frac{\pi }{2}+2\pi n)\cup (\frac{\pi }{2}+2\pi n,\:2\pi +2\pi n)\end{bmatrix}$

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