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Homework Help: Really Hard Trig Equation Someone help please?

  1. May 5, 2008 #1
    1. The problem statement, all variables and given/known data
    Solve the equation:
    cos2x - sinx = (1/2) for [0,2pi).


    2. Relevant equations
    sin^2x + cos^2 = 1


    3. The attempt at a solution

    cos^2(2x) - sin^2x = 1
    1 = 1 Infinite Solutions?
     
  2. jcsd
  3. May 5, 2008 #2

    CompuChip

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    [tex](\cos 2x - \sin x)^2 = \cos^2(2x) - 2 \cos(2x) \sin(x) + \sin^2(x) \neq \cos^2(2x) - \sin^2(x)[/tex]
    and
    [tex]\cos^2(2x) - \sin^2(x) \neq \cos^2(x) + \sin^2(x)[/tex]
    Maybe
    [tex]\cos(2x) = \cos^2(x) - \sin^2(x)[/tex]
    will help?
     
  4. May 5, 2008 #3
    Please rewrite the question. There is repeated misuse of exponents and misplaced variables. Then I'll help all I can!
     
  5. May 5, 2008 #4
    cos2x - sinx = (1/2) for [0,2pi).

    How does that have misplaced exponents and variables? Would Writing parentheses help? Im really sorry for the confusion and my mistake.

    Cos(2x) - sin(x) = (1/2) for [0,2pi).

    Cosine of 2x minus sin of x equals 1 half for the range 0 through 2pie
     
  6. May 5, 2008 #5
    slap yourself.

    Ok, rather than the identity your using ... use:

    [tex]\cos{2x}=1-2\sin^2 x[/tex]

    Just factor and it's solved!!!
     
    Last edited: May 5, 2008
  7. May 5, 2008 #6
    So then

    1 - 2sin^2(x) - sin(x) = (1/2) [-(1/2)]

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

    Quadratic Formula?

    I tried it and got sinx = (1 +/- Sqrroot(5)) / -4)
    x = -54 and 18.
    Somehow that doesn't seem right because radians aren't that big on problems. Any help?
     
  8. May 5, 2008 #7
    -54 and 18 are the answers in degrees, and they are correct
     
  9. Nov 26, 2011 #8
    brendanH the question says between 0 and pi. therefore even if you are right -54 doesnt work
     
  10. Nov 27, 2011 #9

    HallsofIvy

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    And, indeed, since the question itself uses [itex]\pi[/itex], the answer should be in radians, not degrees. There are two solutions between 0 and [itex]\pi[/itex].
     
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