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Math Help - Solving Trig Ids

  1. #1
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    Solving Trig Ids

    I'm given

    cosx (tanx) = 0

    and

    2cos^2x + sin x = 2


    The first one i did...
    cosx ( sinx / cosx ) = 0
    then i got rid of the cosx from sinx / cosx and got...
    cos^2x ( sinx ) = 0
    is that even right? D: Am I just solving for x or what? D:


    for 2cos^2x + sin x = 2 ...
    i moved sinx over to the 2... 2cos^2x = 2-sinx
    then changed it the cos^2 to.. 2( 1-sin^2x) = 2-sinx


    Am i doing these right in the first place..?
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  2. #2
    Super Member craig's Avatar
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    Quote Originally Posted by TeriyakiDonnQ View Post
    cosx (tanx) = 0
    You started right with this.

    cos(x)(tan(x)) = cos(x)(\frac{sin(x)}{cos(x)})

    This is the same as \frac{cos(x)sin(x)}{cos(x)} = sin(x) = 0.

    You then solve this for x.

    For the second one again you had the right idea to start with, let cos^2(x) = 1 - sin^2(x), your equation is now:

    2(1 - sin^2(x)) + sin(x)-2 = 0

    Rearranging this gives us sin^2(x) - \frac{1}{2}sin(x) = 0

    If we now replace sin(x) with S, we get a regular quadratic in S to solve:

    S^2 - \frac{1}{2}S = 0
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  3. #3
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    ohh~ okayy
    i get it thank youuu!
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