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Help with a problem required =D

  1. May 7, 2008 #1
    cos(x)sin(x) - .25cos(x) = 0 for 0° ≤ x ≤ 360°

    We had a problem thrown up today just before we left and asked to attempt it tonight. But I have been starring at the problem for over an hour now, and all I've gotten is being able to plug it into the calculator and finding the values of X when Y = 0, lol and that was when I started attempting it..

    I'm completely lost and the next section of the book doesn't really help me out. I know I'm missing something completely obvious, but I'm brainfarting bad.

    I was assuming it had something to do with Sum and difference formulas for sine and cosine which is the next section in the book, but the more I look at it, the harder my head is pounding, and honestly I'm just about to give up on it and bring in the calculator with the answers. lol.

    Thanks for any help you guys can give..
  2. jcsd
  3. May 7, 2008 #2
    This is actually easier than you think. How might the distributive property help?
  4. May 7, 2008 #3
    Sorry for the other posts, firefox or the internet or something sent the post response thing more than once.. Oh well.. lol
  5. May 7, 2008 #4


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    I think I got the right reply moved into the right thread for you since the other one got a response before you edited.
  6. May 7, 2008 #5
    Yup, thanks.
  7. May 10, 2008 #6


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    Now, Tedjn's point is that sin(x)cos(x)- 0.25 cos(x)= 0 is the same as cos(x)(sin(x)- 1)= 0. What does that tell you?
  8. May 10, 2008 #7
    uhm you mean cos(x)(sin(x)- 0,25)= 0, right?
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