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Math Help - Trig Equation

  1. #1
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    Trig Equation

    3sin(2A) = sin(A)

    There are a number problems like this in my textbook and I am able to solve them, but I always end up with a polynomial in some trig function. I wanted to know if there were any less time consuming approaches to this question?


    Here's what I did for the above example:


    3sin(2A) = sin(A)

    6sin(A)cos(A) = sin(A)

    6(1-cos^2(A))(cos(A)) = 1 - cos^2(A)

    6cos^3(A) - cos^2(A) - 6cos(A) + 1 = 0

    and then solved to find cos(A).


    I have a feeling that I am making the question more time-consuming than neccessary - is there a simpler approach?
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  2. #2
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    Re: Trig Equation

    Hello, kinhew93!

    Your substitution is incorrect: . \sin A \:=\:\sqrt{1-\cos^2\!A}


    3\sin2A \:=\: \sin A

    I would solve it like this . . .

    n . . . . . . . . . 6\sin A\cos A \:=\:\sin A

    . . . . . 6\sin A\cos A - \sin A \:=\:0

    Factor: . \sin A(6\cos A - 1) \:=\:0


    . . \sin A \:=\:0 \quad\Rightarrow\quad \boxed{A \:=\:\pi n}

    . . 6\cos A - 1 \:=\:0 \quad\Rightarrow\quad \cos A \:=\:\tfrac{1}{6} \quad\Rightarrow\quad \boxed{A \:=\:\cos^{-1}\left(\tfrac{1}{6}\right) + 2\pi n}
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  3. #3
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    Re: Trig Equation

    Quote Originally Posted by Soroban View Post
    Hello, kinhew93!

    Your substitution is incorrect:
    In this case it is because you can times through by sinA...

    Thanks though you were a great help! I knew I was make it far to complicated!
    Last edited by kinhew93; January 1st 2013 at 02:39 PM.
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