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Solving Trigonometric Equations

  1. Nov 16, 2011 #1
    1. The problem statement, all variables and given/known data

    [itex]2sin\Theta-\sqrt{2}=0[/itex]



    2. Relevant equations



    3. The attempt at a solution

    [itex]2sin\Theta=\sqrt{2}[/itex]

    [itex]\frac{2sin\Theta}{2}=\frac{\sqrt{2}}{2}[/itex]

    So here's my problem. Should it be positive or negative? Or just positive? Since the square root of any positive number can be positive or negative, right? Is there a possibility that the answer can be both positive and negative?

    [itex]sin\Theta=\pm\frac{\sqrt{2}}{2}[/itex]

    if so, [itex]\Theta=45 degrees[/itex] [itex] or[/itex] [itex] 135 degrees[/itex] ---- for positive value of [itex]sin\Theta[/itex]

    [itex]\Theta=-45 degrees[/itex] [itex] or [/itex] [itex]315 degrees[/itex] [itex] or[/itex] [itex] 225 degrees[/itex] ---- for negative value of [itex]sin\Theta[/itex]

    So, I'm confused if:

    [itex]2sin\Theta-\sqrt{2}=0[/itex]

    [itex]2sin\Theta=\sqrt{2}[/itex]

    [itex]\frac{2sin\Theta}{2}=\frac{\sqrt{2}}{2}[/itex]

    Can be:


    [itex]2sin\Theta-(\pm\sqrt{2})=0[/itex]

    [itex]2sin\Theta=(\pm\sqrt{2})[/itex]

    [itex]\frac{2sin\Theta}{2}=(\pm\frac{\sqrt{2}}{2})[/itex]

    So can it be?
    Is it mathematically correct to think of it that way? Sorry if this is such a stu*id question. :)
    Hope you can help me! :)
     
  2. jcsd
  3. Nov 16, 2011 #2

    danago

    User Avatar
    Gold Member

    [itex]\sqrt{2}[/itex] is positive and separate from its negative counterpart, [itex]-\sqrt{2}[/itex]. If the question did not say [itex]± \sqrt{2}[/itex], then you can safely just worry about the positive value.

    If, on the other hand, the question was [itex]4sin^2 \Theta-2=0[/itex], then when you take the square root you must be sure to include the ±.

    It is sort of like saying [itex]x=\sqrt{25}[/itex] is equal to positive 5 only, however if i said that [itex]x^2=25[/itex], then x could be +5 or -5.
     
  4. Nov 16, 2011 #3
    Thank you for clearing this out for me!
    Sadly that was on my quiz and I totally included the negative root as a solution for the equation. :(

    At least now I know it. Thanks for answering! :)
     
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