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Thread: Algebraic Trig Solutions

  1. #1
    ATC
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    Algebraic Trig Solutions

    "Find all solutions to $\displaystyle 15sec(2x)sin(2x) + 9sec(2x) = 10sin(2x) + 6$ algebraically."

    I am having trouble getting very far with this. Thanks.

    Edit: I have tried messing around with the double argument formulas and turning them into different ratios.
    Last edited by ATC; Nov 21st 2010 at 03:43 PM.
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  2. #2
    Master Of Puppets
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    Here's a start

    $\displaystyle \sec(2x)\sin(2x) = \frac{\sin(2x)}{\cos(2x)} = \tan(2x)$
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  3. #3
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    Hello, ATC!

    Find all solutions to: .$\displaystyle 15\sec2x\sin2x + 9\sec2x \:=\: 10\sin2x + 6$

    Edit: I have tried messing around with the double argument formulas.
    Since they are all double angles, why bother?

    Multiply through by $\displaystyle \cos2x\!:$

    . . $\displaystyle 15\sin2x + 9 \;=\;10\sin2x\cos2x + 6\cos2x$

    . . $\displaystyle 10\sin2x\cos2x - 15\sin2x + 6\cos2x - 9 \;=\;0$

    Factor: .$\displaystyle 5\sin2x\bigg[2\cos2x - 3\bigg] + 3\bigg[\2\cos2x - 3 \bigg] \;=\;0$

    Factor: .$\displaystyle (2\cos2x - 3)(5\sin2x + 3) \;=\;0$


    And we have:

    . . $\displaystyle 2\cos2x - 3 \:=\:0 \quad\Rightarrow\quad \cos 2x \:=\:\frac{3}{2} \quad\hdots\;\text{ No real solutions.}$

    . . $\displaystyle 5\sin2x + 3 \:=\:0 \quad\Rightarrow\quad \sin2x \:=\:-\frac{3}{5}$

    . . $\displaystyle 2x \;=\;\arcsin(-0.6) \;=\; \begin{Bmatrix}\text{-}36.9^o + 360^on \\ 116.9^o + 360^on \end{Bmatrix}$

    Therefore: .$\displaystyle x \;=\;\begin{Bmatrix}\text{-}18.45^o + 180^on \\ 58.45^o + 180^on \end{Bmatrix}\,\text{ for some integer }n.$

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  4. #4
    ATC
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    Thank you...multiplying by cos2x made it so much simpler.
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