# Solving Sin(x)=0

1. May 19, 2015

### Scheuerf

How would I solve for sin(x)=0? When I enter arcsin(0) into a calculator, I get 0, but there should be multiple solutions if I'm not wrong. How would I find others, all between 0 and 720 for example?

2. May 19, 2015

### kontejnjer

Since the sine function is periodic, you can figure it out by induction: on the interval $(0,2\pi)$, it has two zeroes, namely $\{0,\pi\}$. Since the period $T$ is $2\pi$, and $\sin(x+nT)=\sin(x)$ is valid for $n\in ℤ$, the complete solution set can be seen to be $x=n\pi;n\in ℤ$.

3. May 19, 2015

### Staff: Mentor

Using arcsin doesn't get you far. By common agreement, the arcsine function is the inverse of the Sin() function, which is the same as the sin() function, but with a domain restricted to $[-\pi/2, \pi/2]$. This restriction makes Sin() a one-to-one function, therefore a function that has an inverse. The restricted domain of $[-\pi/2, \pi/2]$ for Sin is the range of its inverse, arcsin. Taking arcsin(0) will get you only one value; namely, 0.

To find all solutions of the equation sin(x) = 0 you have to understand the periodicity of the sin function and that its intercepts are all of the integer multiples of $\pi$.

4. May 19, 2015

### Drafter

You can the general solution formula of......2kπ. Where k is any integer.
Here is a the general formula of cosθ=1/2
I use the term "general formula" loosely since I do not know it's exact terminology, but anyways, here is cos(θ)=1/2.

Assuming you have a solid understanding of the unit circle, you can easily have a go at applying this general formula to the sin(θ)=0 with no problem.