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Solving Sin(x)=0

  1. May 19, 2015 #1
    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. jcsd
  3. May 19, 2015 #2
    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].
     
  4. May 19, 2015 #3

    Mark44

    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##.
     
  5. May 19, 2015 #4
    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.
    3eGYLRz.png


    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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