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Math Help - Solving a trig equation

  1. #1
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    Solving a trig equation

    Hi,

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

    2sin(x)cos(x) + sin(x) = 0

    but i get as far as

    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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    e^(i*pi)'s Avatar
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    Quote Originally Posted by aceband View Post
    Hi,

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

    2sin(x)cos(x) + sin(x) = 0

    but i get as far as

    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

    sin(x)(2cos(x)+1) = 0

    Either sin(x)=0 or 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:

    2sin(x)cos(x) + sin(x) = 0

    but i get as far as

    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:

    \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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