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Thread: Factorable Trig equations (I'm totally lost with this one)

  1. #1
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    Factorable Trig equations (I'm totally lost with this one)

    Okay.. I apologize for my inability to use the fancy math graphics. I'm going to finally put some time into learning that soon.

    ---

    Solve: 2 sec^2 (-)/2 + 3 sec (-)/2 - 2 = 0

    provided 0 deg < (-) < 360 deg

    ---

    I've already tried a couple (most likely completely wrong) approaches... If someone could just walk me through this, I'd really appreciate it.
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  2. #2
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    Quote Originally Posted by Savior_Self View Post
    Okay.. I apologize for my inability to use the fancy math graphics. I'm going to finally put some time into learning that soon.

    ---

    Solve: 2 sec^2 (-)/2 + 3 sec (-)/2 - 2 = 0

    provided 0 deg < (-) < 360 deg

    ---

    I've already tried a couple (most likely completely wrong) approaches... If someone could just walk me through this, I'd really appreciate it.
    Do you mean $\displaystyle sec^2(\frac{\theta}{2})$?

    This is a quadratic equation in $\displaystyle sec(\frac{\theta}{2})$. To make it easier substitute $\displaystyle u = sec(\frac{\theta}{2})$ to give

    $\displaystyle 2u^2 + 3u - 2 = 0$

    (2u-1)(u+2) = 0 so u = 0.5 or u = -2

    ie: $\displaystyle sec(\frac{\theta}{2}) = 0.5$ in which case we can take the reciprocal of both sides to give us $\displaystyle cos(\frac{\theta}{2}) =2$. You'll get an error if you try solving so this is not a real solution.

    $\displaystyle sec(\frac{\theta}{2}) = -2$

    Again take the reciprocal to give $\displaystyle cos(\frac{\theta}{2}) = -0.5$
    this gives $\displaystyle \frac{\theta}{2} = 120^o$ and $\displaystyle \theta = 240^o$

    This is the only solution if I've plotted my graph right
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  3. #3
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    Quote Originally Posted by Savior_Self View Post
    Okay.. I apologize for my inability to use the fancy math graphics. I'm going to finally put some time into learning that soon.

    ---

    Solve: 2 sec^2 (-)/2 + 3 sec (-)/2 - 2 = 0

    provided 0 deg < (-) < 360 deg

    ---

    I've already tried a couple (most likely completely wrong) approaches... If someone could just walk me through this, I'd really appreciate it.
    $\displaystyle 2\sec^2{\frac{x}{2}} + 3\sec{\frac{x}{2}} - 2 = 0$

    Let $\displaystyle X = \sec{\frac{x}{2}}$ so that the equation becomes

    $\displaystyle 2X^2 + 3X - 2 = 0$

    $\displaystyle 2X^2 + 4X - X - 2 = 0$

    $\displaystyle 2X(X + 2) - (X + 2) = 0$

    $\displaystyle (X + 2)(2X - 1) = 0$

    $\displaystyle X + 2 = 0$ or $\displaystyle 2X - 1 = 0$

    $\displaystyle X = -2$ or $\displaystyle X = \frac{1}{2}$

    $\displaystyle \sec{\frac{x}{2}} = -2$ or $\displaystyle \sec{\frac{x}{2}} = \frac{1}{2}$.

    Can you solve for $\displaystyle x$?
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  4. #4
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    Thanks a ton to both of you. Turns out I was actually somewhat close, I just missed some critical steps that you've showed me. Thanks!
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