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Trig quick question

  1. Aug 21, 2012 #1
    Solve for values of θ in the interval 0 ≤ θ ≤ 360,

    2sinθ = cosecθ

    now if I do:

    2sinθ = 1/(sinθ) and multiply by sinθ

    I can solve sinθ = ±√(1/2)

    but if I solve 2sinθ - 1/(sinθ) = 0 and factorise to get sinθ(2 - (1/sin^2θ)) = 0

    I get sinθ = 0 which gives me more solutions then needed. I've drawn the graphs of both functions and they meet where the solutions of θ are such that sinθ = ±√(1/2) , then why do I get more solutions if I factorise?
     
  2. jcsd
  3. Aug 21, 2012 #2

    Mentallic

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    Homework Helper

    Because at [itex]\sin\theta=0[/itex] the other factor is undefined.

    Think about the function [tex]y=\frac{x(x+1)}{x}[/tex] this is pretty much equivalent to [itex]y=x+1[/itex] except for the fact that it has a hole at x=0 (the coordinate (0,1)). If we solved this equation for when y=0, clearly the only answer is x=-1, but if we factored out x then what we'd have is:

    [tex]x\left(\frac{x+1}{x}\right)=0[/tex]

    Which seems like it would have an x=0 solution, but it doesn't by the same reasoning.
     
  4. Aug 21, 2012 #3
    I see, thanks.

    But why does this happen? Is there a particular reason or just one of those things.
     
  5. Aug 21, 2012 #4

    Mentallic

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    It happens because when we use the rule that if [itex]ab=0[/itex] then either a, b or both are equal to zero, we are assuming that a and b are real numbers. Undefined numbers and infinite are not real.

    Without using any rigor, if we use an undefined number like 1/0 such that a=0 and b=1/0, then ab=1 (again, I'm abusing the maths here just to explain a point, don't take it as being correct). As you can see while we solved for a=0, it turns out that ab didn't turn out to be equal to zero, hence we cannot take a=0 because it is not a solution.
     
  6. Aug 21, 2012 #5
    thank you
     
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