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Thread: Solving a trig equation

  1. #1
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    Solving a trig equation

    Hi,

    The question is solve for ipi =< x <= pi:

    $\displaystyle 2sin(x)cos(x) + sin(x) = 0$

    but i get as far as

    $\displaystyle 2cos^{3}(x) + cos^{2}(x) - 2cos(x) - 1 = 0$

    and then can't get further. Am i on the right track?
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  2. #2
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    Quote Originally Posted by aceband View Post
    Hi,

    The question is solve for ipi =< x <= pi:

    $\displaystyle 2sin(x)cos(x) + sin(x) = 0$

    but i get as far as

    $\displaystyle 2cos^{3}(x) + cos^{2}(x) - 2cos(x) - 1 = 0$

    and then can't get further. Am i on the right track?
    It may work but not the way I'd go. I'd factorise the equation

    $\displaystyle sin(x)(2cos(x)+1) = 0$

    Either $\displaystyle sin(x)=0$ or $\displaystyle cos(x) = -\frac{1}{2}$
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  3. #3
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    Quote Originally Posted by aceband View Post
    Hi,

    The question is solve for ipi =< x <= pi:

    $\displaystyle 2sin(x)cos(x) + sin(x) = 0$

    but i get as far as

    $\displaystyle 2cos^{3}(x) + cos^{2}(x) - 2cos(x) - 1 = 0$

    and then can't get further. Am i on the right track?
    No.

    You're meant to factorise:

    $\displaystyle \sin x (2 \cos x + 1) = 0$

    Then use the null factor law to get two simple equations that are easily solved.
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