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Sin(ax+b)=c, solve for x

  1. Jun 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Hello,

    I have a homework problem where I am trying to solve for x in the following equation:

    sin(ax+b)=c


    2. Relevant equations



    3. The attempt at a solution

    This is my answer:

    x=[itex]\frac{arcsin(c)-b}{a}[/itex]

    My question is, is this all I can do? Unless there is something I am neglecting, it seems fairly restrictive on what x can equal, due to the domain restrictions of the arcsin. Is there any other way of solving this without using arcsin?
     
  2. jcsd
  3. Jun 2, 2012 #2

    SammyS

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    Yes, there are other answers.

    If sin(ax+b)=c then also sin(ax+b+2πk)=c where k is an integer. This is because of the sine function's periodicity.

    It's also true that sin(π-θ) = sin(θ) . Therefore, if sin(ax+b)=c, then also sin(π-ax-b)=c .
     
  4. Jun 2, 2012 #3

    Curious3141

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    No you need the arcsin(c). However, that just gives you the principal value (the single value between -π/2 and π/2). To get the complete solution set, you need to consider the periodicity of the sine function. Since the sine function is periodic with period 2π, your complete solution set is described by:

    [itex]x = \frac{\arcsin c - b + 2k\pi}{a}[/itex] where [itex]c \neq 0[/itex]

    and

    [itex]x = \frac{k\pi - b}{a}[/itex] where [itex]c = 0[/itex]

    since in the latter case, the sines of all multiples of π equal zero.

    As SammyS pointed out, [itex]\sin(\pi - \theta) = \sin \theta[/itex]. In fact, this is true for all odd multiples of [itex]\pi[/itex], so an additional solution exists:

    [itex]x = \frac{(2k+1)\pi - \arcsin c - b}{a}[/itex] for any c.

    In all of the above, k can take any integer value (positive, negative or zero).

    If you're asked to restrict your solutions to a narrow range, just use the values that satisfy your range.
     
    Last edited: Jun 2, 2012
  5. Jun 4, 2012 #4
    Thanks a lot, folks. I really appreciate it. I figured it had something to do with k[itex]\pi[/itex] and 2k[itex]\pi[/itex], so I'll mull over these posts some more.
     
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