How do you find the domain of a fuction such as f(x)=(x+sin x)/cos x?
since denominator, cos x is zero when $\displaystyle x= \frac{n \pi}{2}, n\in \mathbb{I}$
So, domain is all real values except this value.
Domain = { $\displaystyle x\in \mathbb{R}/ x\ne\frac{n \pi}{2}, n\in \mathbb{I}$}
or you can write this way:
Domain $\displaystyle = \mathbb{R} - \frac{n \pi}{2}, n\in \mathbb{I}$
For what x between $\displaystyle -\frac{3 \pi}{2}$ and $\displaystyle \frac{3 \pi}{2}$ is $\displaystyle \cos(x)=0$ ?
The formula is as follows :
$\displaystyle \cos(a)=\cos(b) \implies \text{either } \left\{\begin{array}{ll} a=b+2k \pi \\ a=-b+2k' \pi \end{array} \right. \text{ where k and k' } \in \mathbb{Z}$
$\displaystyle \cos \left(\tfrac \pi 2\right)=0$ so in fact, it will give $\displaystyle \cos(x)=0 \Longleftrightarrow x=\frac \pi 2+k \pi$ (which is o_O's formula)
This will give you all the possibilities for $\displaystyle -\frac{3 \pi}{2} \le x \le \frac{3 \pi}{2}$, that is... try to fit values for k such that x remains between these two values.
Oh, yes, I realised it later. There was a mistake. Actually it should be,
Domain={$\displaystyle x\in \mathbb{R}/ x\ne\frac{(2n-1) \pi}{2}, n\in \mathbb{I}$}
or you can say
Domain=$\displaystyle \mathbb{R} - \frac{(2n-1) \pi}{2}, n\in \mathbb{I}$
because, cos x is zero for odd multiples of $\displaystyle \frac{\pi}{2}$
i know this thread is kinda done with, but cant the domain really just be any real number? in algebra we know that the domain of a function can be any real number, so long as it has only one range, but thats besides the point of the fact that shouldnt the answer be any real number?