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Thread: need factors part of trig eqation

  1. #1
    Member sinewave85's Avatar
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    need factors part of trig eqation

    Ok, working on a trig equation, have part of it down to:

    $\displaystyle -4\cos^{2}x+2\sqrt{1-\cos^{2}x}+3=0$

    I need the left side factored (so I can use the zero rule to find values for x) but I can't seem to get it. Can anyone else?
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  2. #2
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    Quote Originally Posted by sinewave85 View Post
    Ok, working on a trig equation, have part of it down to:

    $\displaystyle -4\cos^{2}x+2\sqrt{1-\cos^{2}x}+3=0$

    I need the left side factored (so I can use the zero rule to find values for x) but I can't seem to get it. Can anyone else?
    Note, the error $\displaystyle \sqrt{1-\cos^2 x}=\sin x$
    In fact it is, $\displaystyle \sqrt{1-\cos^2 x}=|\sin x|$
    The presense of the absolute value function overcomplicates things.

    What you can do, is express it as, (add and subtract 1)
    $\displaystyle -4\cos^2 x+4+2\sqrt{1-\cos^2 x}-1=0$
    Thus,
    $\displaystyle 4(1-\cos^2 x)+2\sqrt{1-\cos^2 x}-1=0$
    If,
    $\displaystyle \sqrt{1-\cos^2 x}=y$
    Then,
    $\displaystyle 4y^2+2y-1=0$
    You have a quadradic equation.

    Just remember to check solutions because it is necessary though not suffienct.
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  3. #3
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    Hello, sinewave85!

    Have you ever tried to factor something with a square root in it?


    $\displaystyle -4\cos^2\!x + 2\sqrt{1-\cos^2x} + 3 \;=\;0$

    Remember "radical equations" from Algebra 2?

    We have: .$\displaystyle 2\sqrt{1-\cos^2\!x} \;=\;4\cos^2\!x - 3$

    Square both sides: .$\displaystyle 4\left(1 - \cos^2\!x\right) \;= \; 16\cos^4\!x - 24\cos^2\!x + 9$

    . . and we have: .$\displaystyle 16\cos^4\!x - 20\cos^2\!x + 5\;=\;0$

    Quadratic Formula: .$\displaystyle \cos^2\!x \;= \;\frac{-(-20) \pm \sqrt{20^2 - 4(16)(5)}}{2(16)} \;=\;\frac{20 \pm\sqrt{80}}{32}$

    . . and we have: .$\displaystyle \cos^2\!x \;=\;\frac{5 \pm\sqrt{5}}{8}\quad\Rightarrow\quad\boxed{\cos x\;=\;\pm\sqrt{\frac{5\pm\sqrt{5}}{8}}} $

    Don't forget to check for extraneous roots.


    [And you wonder why you couldn't factor it . . . ]
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  4. #4
    Member sinewave85's Avatar
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    Thanks for the help -- both of you! I got it worked out, and I got the right answers, along with a few false ones. Just in case you are currious, the original problem was sin2x = cos3x, and the answers for x between 0 and 180 degrees are 18, 90, and 160 degrees. I am not sure I solved it the best possible way, but I did get the right answers -- with a little help . Thanks again for getting me through my brain fart.
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