Solving Sin(x)=0

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

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