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Simple trig identification equation question which i can not answer

  1. May 11, 2007 #1
    1. The problem statement, all variables and given/known data

    Solve for X

    2. Relevant equations

    2sin^2x - cosX - 1 = 0

    3. The attempt at a solution

    I have gone in this order

    I turned 2sin^2x into 1 - cos2x

    so
    1 - cos2x - cosx -1 = 0

    This turns into
    Cos2x + cosx = 0


    I dont know what to do next...
    could someone plz help :(, much appreciated
     
  2. jcsd
  3. May 11, 2007 #2
    I'm not sure if you mean 2 times sin x squared or 2 times sin 2x?
    But it if is former, then...
    2(sin x)^2 - cos x - 1 = 0
    2(1 - [cos x]^2) - cos x - 1 = 0
    2 - 2[cos x]^2 - cos x - 1 = 0
    -2[cos x]^2 - cos x + 1 = 0
    from here you just treat it as you would a quadratic equation.
     
  4. May 11, 2007 #3
    Follow up....

    Following on Kuno's advice, you have:

    -2 cos^2 x - cos x + 1 = 0

    Which is really like asking:

    -2x^2 - x + 1 = 0

    In other words: (-2x + 1)(x+1) = 0

    x = -1, x=1/2

    Thus you are really asking, when does :

    cos x = -1 ?

    and when does

    cos x = 1/2 ?

    Which happens when

    x = pi, x = pi/3 or if you prefer degrees: x = 180, 60

    Which is the answer to your problem. (Assuming of course that you wanted to write 2 * sin^2 x and not 2*sin 2x)
     
  5. May 11, 2007 #4
    thank you, thank you, thank you, thank you, thank you SO MUCH!.


    I relise what i did wrong, i was using the wrong identity, you know how
    cos2a = 2sinx^2 -1

    I was using that identity instead of the sin^2x + cos^2x = 1 identity.

    Just out of curiousity, say i used that first identity and got the answer
    cos2x +cosx = 0
    Is it possible to solve a equation like that or not ?
     
  6. May 11, 2007 #5
    Actually that should read:

    cos 2x = 1 - 2 sin^2 x

    This second identity allows you to say also:

    1 - 2*sin^2 x = 1 - 2*(1 - cos^2x) and thus:

    cos 2x = 2*cos^2 x - 1

    I guess you really mean to ask: Does the equation :

    cos 2x + cos x = 0

    Have any solutions? (assuming for 0..x..2pi)

    Well, we can start with this identity (the double angle formula):

    cos 2x = 2*cos^2 x - 1

    We have the following:

    2*cos^2 x + cos x - 1 = 0

    Which is like the following:

    2u^2 + u - 1 = 0

    or (u+1)(2u-1)=0 that is, u = -1, u=1/2

    Thus our question really means :

    When does u (that is cos x) = -1,1/2 or:
    when does cos x = -1, cos x = 1/2 ?

    Thus x = pi, pi/3

    You will find that many many equations of this sort are solved in this manner - you are essentially doing what we call "substitution" of a function, something very common in calculus - reducing a problem to a format which has a known solution. In this case you are reducing trigonometric functions to a polynomial, solving the polynomial and then finding when the trigonometric function has those values.

    Here is also a good page for Trigonometric identities:

    http://www.sosmath.com/trig/Trig5/trig5/trig5.html

    JJ +
     
  7. May 22, 2007 #6

    disregardthat

    User Avatar
    Science Advisor

    [tex]\cos^2{x} + \cos{x} = 0[/tex]

    substitute with x:

    [tex]x^2 + x = 0[/tex]

    [tex]\frac{-1 \pm 1}{2}[/tex]

    [tex]x_1 = -1[/tex]
    [tex]x_2 = 0[/tex]

    [tex]\cos{x} = -1[/tex]
    [tex]\arccos{-1} = \pi[/tex]

    [tex]\cos{x} = 0[/tex]
    [tex]\arccos{0} = \frac{\pi}{2}[/tex] and [tex]\frac{\pi}{2} + \pi = \frac{3\pi}{2}[/tex]

    [tex]x = \frac{\pi}{2} , \pi , \frac{3\pi}{2}[/tex]
     
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