When \[\sin x=0\] what does x equal?

Hint: In order to determine the solution, first of all we need to know the value of sin in such a way that expression should be 0. But as we know that sin is also known as a periodic function that oscillates at a regular interval that is 0. At \[x=0,\,\pi ,\,2\pi \] in the domain.

Complete step-by-step solution:
We have given the trigonometry equation \[\sin x=0\]. In this equation, we have to find the values of x.
The given expression is \[\sin x=0\]
If you see in this expression then sin is also known as periodic function that oscillates at regular interval and it crosses the x axis (that is 0) at \[x=0,\,\pi \] and \[2\pi \] in the domain \[\left[ 0,2\pi\right] \] and it continues to cross the x axis at every integer multiple of \[\pi \]
For more understanding clearly figure is given below:

Range of this above graph is given below that is
Graph \[\left( \sin x\left[ -1,1 \right] \right)\]
So whenever \[\sin x=0\] , we have that
\[x=\pi \pm k\pi \] For all values of k in the set of integers.
That means we can consider the value of k as if \[k=0,1,2,.........N\] where N is some arbitrary large integer.
Then \[\sin x=0\] for \[x=0,\pm \pi ,\pm 2\pi ,......\pm 2N\pi \]

Note: One must be careful while taking values from the trigonometry table and cross check at least once to avoid error in the answer. If you put a multiple of \[2\pi \] that is \[2N\pi \] where N is a larger integer value then also the expression becomes 0. In this question, students can also check the values of x in the equation given in the question and if the solution comes as 0 then, the values of x are correct.