Solving $\sin x = 0$, I get $x=k\times 360^\circ$ and $x=k\times 360^\circ+180^\circ$, but the answer is $x=k\times 180^\circ$.

Question

How to prove that $\sin x = 0$ when $x = k\times180^\circ$?

I tried doing it, but I got these results: $$x=k\times 360^\circ \qquad x=k\times 360^\circ+180^\circ$$

Answer 1

$$\sin x = 0$$

For $n \in \mathbb{Z}$, $x = \pi + 2\pi n$ OR $x = 2\pi n$.

$$x = \pi + 2\pi n \implies x = …, -5\pi, -3\pi, -\pi,\pi, 3\pi, 5\pi, …$$

$$x = 2\pi n \implies x = …, -6\pi, -4\pi, -2\pi, 0, 2\pi, 4\pi, 6\pi, …$$

If you noticed, combining them would give all integer multiples of $\pi$, so you just combine them. $$x = \pi n$$

You could demonstrate the same idea in degrees: $x = 180+360n$ and $x = 360n$ can be combined to give $x = 180n$.