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Thread: Simple trig equation

  1. #1
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    Simple trig equation

    5sin(A)cos(A) - sin(A) = 0 (A is between 0 and 360 degrees)

    by dividing through by sin(A) I get:

    cos(A) = 1/5

    so A = 78.5, 281.5

    Though clearly A = 0 also satisfies this equation so could somebody please explain why I have missed it in my method?
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  2. #2
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    Re: Simple trig equation

    by dividing, you "lost" solutions. factoring is the way to go here ...

    $\displaystyle 5\sin(A)\cos(A) - \sin(A) = 0$

    $\displaystyle \sin(A)[5\cos(A) - 1] = 0$

    $\displaystyle \sin(A) = 0 \implies A = 0 \, , \, 180 \, , \, 360$

    $\displaystyle \cos(A) = \frac{1}{5} \implies A = \arccos\left(\frac{1}{5}\right) \, , \, 360 - \arccos\left(\frac{1}{5}\right)$
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  3. #3
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    Re: Simple trig equation

    Thanks but I still don't understand how those solutions were lost? If attempting a question like this how will I know that dividing will lose solutions? Are you able to explain this in a way that I can understand?

    Thanks
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  4. #4
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    Re: Simple trig equation

    Has no one ever explained to you that you cannot divide by 0? Either sin(A) is not 0 so you can divide by it or sin(A)= 0.

    A better way of doing this problem would be to write 5 sin(A)cos(A)- sin(A)= sin(A)(5 cos(A)- 1)= 0. The product of two numbers is 0 if and only if one or the other of them is 0: either sin(A)= 0 or 5cos(A)- 1= 0.
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  5. #5
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    Re: Simple trig equation

    Hello, kinhew93!

    Thanks, but I still don't understand how those solutions were lost.

    You will lose a solution if you divide by the variable
    . . (or an expression containing the variable).


    Suppose we have the equation: .$\displaystyle x^2 - 3x \:=\:0$ .[1]

    It is obviously a quadratic
    . . and you know it has two solutions, right?

    The proper procedure is:

    . . Factor: .$\displaystyle x(x-3) \:=\:0$
    . . Set each factor equal to zero and solve: .$\displaystyle \begin{Bmatrix}\boxed{x \:=\:0} \\ x-3\:=\:0 & \Rightarrow& \boxed{x \:=\:3} \end{Bmatrix}$


    But suppose you decide to divide [1] by $\displaystyle x\!:\;\;\frac{x^2}{x} - \frac{3x}{x} \:=\:\frac{0}{x}$
    Then you have: .$\displaystyle x - 3 \:=\:0$
    . . which has only one solution: .$\displaystyle x = 3$

    See?
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  6. #6
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    Re: Simple trig equation

    Seems so obvious now... thanks
    Last edited by kinhew93; Dec 28th 2012 at 03:06 PM.
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