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anonymous

  • one year ago

how many solutions in the inteval 0<x<2pie to the following of (2xsin3x-1)(cos2x+1)=0

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  1. zepdrix
    • one year ago
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    Hey, start by applying your `Zero Factor Property`,\[\large\rm 2x \sin3x-1=0\qquad\qquad\qquad \cos2x+1=0\]

  2. zepdrix
    • one year ago
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    Are you sure that first one has an \(\large\rm x\) and \(\large\rm sin3x\) in it? :o Hmm that's going to be tricky..

  3. anonymous
    • one year ago
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    Yes

  4. zepdrix
    • one year ago
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    The second one shouldn't be too bad. Subtract 1 from each side,\[\large\rm \cos(\color{orangered}{2x})=-1\]Recall that cosine is -1 at an angle of pi.\[\large\rm \color{orangered}{2x=\pi+2k \pi}\]Any number of full spins will get us back to that same point, so we add an amount of 2pi's on the end like that. Solving for x,\[\large\rm x=\frac{\pi}{2}+k \pi\]For k=0, we get \(\large\rm x=\frac{\pi}{2}\) For k=1, we get \(\large\rm x=\frac{3\pi}{2}\) Any other values of k will take us outside of our interval. So that's all the solutions we're getting from the second bracket.

  5. zepdrix
    • one year ago
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    Hopefully that makes sense. Hmm I'm not sure what to do about the other bracket though :p

  6. zepdrix
    • one year ago
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    Oh, it simply says "how many", not to actually find them... oh oh interesting.

  7. zepdrix
    • one year ago
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    Is this for calculus? :)

  8. zepdrix
    • one year ago
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    cause maybe we could count the critical points.

  9. zepdrix
    • one year ago
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    hmm

  10. anonymous
    • one year ago
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    Thanks for your help and Yes it is for calculus. :)

  11. zepdrix
    • one year ago
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    Let's see if any of these smart guys have an idea :d @ganeshie8 @Kainui @dan815

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