What is the domain of `f(x)=cscx`?

The question gets much easier if we recall that, by definition, the function you're dealing with is `\frac{1}{\sin(x)}`.

Being a fraction, we have to make sure that the denominator is non-zero; and since there are no roots or logarithms, it is the only thing we have to study.

The domain of the function will thus be the following set: `{ x \in \mathbb{R} | \sin(x)\ne0}`

The sine function is defined as the projection of a point on the unit circle on the `y` axis, and thus `\sin(x)=0` if and only if the point belongs to the `x` axis.

The only two points which are on both the unit circle and on the `x` axis are `(1,0)` and `(-1,0)`, and they are given by a rotation of 0 and `\pi` radiants. Because of the periodicity of the sine function, 0 radiants is the same as `2k\pi` radians, and `pi` radians are the same as `(2k+1)\pi` radians, for each `k \in \mathbb{Z}`.

Finally, our answer is ready: we need to exclude from the domain all the points of the form `2k\pi`, for each `k \in \mathbb{Z}`. Using a proper notation, the domain is the set `D_f={x \in \mathbb{R} | x \ne 2k\pi, \forall k \in \mathbb{Z}`}