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marasofia1616 Group Title

Find all solutions to the equation. cos2x + 2 cos x + 1 = 0

  • one year ago
  • one year ago

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  1. rsnoob Group Title
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    For cos(2x) * (2cos(x) + 1) = 0, use the double angle identity for cos(2x), which is cos^2 x - sin^2 x = cos^2 x - (1-cos^2) = 2cos^2 x - 1. So we have (2cos^2 x - 1)(2cos x + 1) = 0. So 2cos^2 x -1 = 0 or x = 0 and 2pi. For 2sec^2 x + tan^2 x - 3 = 0, use the identity sec^2 x = tan^2 x + 1, so we have 2(tan^2 x + 1) + tan^2 x - 3 = 0 or 2tan^2 x + tan^2 x - 1 = 0 or 3 tan^2 x = 1. So x = pi/2, pi/2 + pi = 3pi/2.

    • one year ago
  2. marasofia1616 Group Title
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    hi the first thing I tried to solve this problem was to factor it so I have (cos x+1)(cos x+1) but now I dont know what to next

    • one year ago
  3. tkhunny Group Title
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    Unclear how to proceed. What is the first term, \(\cos(2x)\) or \(\cos^{2}(x)\)?

    • one year ago
  4. marasofia1616 Group Title
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    cos \[\cos^{2}x\]

    • one year ago
  5. tkhunny Group Title
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    Perfect. The we have only to solve - by factoring. \(\cos^{2}x + 2\cos x + 1 = 0\) \((\cos(x) + 1)^{2} = 0\) Now what?

    • one year ago
  6. marasofia1616 Group Title
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    That is where I got stuck I am not sure what to do after factoring

    • one year ago
  7. tkhunny Group Title
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    What's the point of factoring? Why do we do that at all? If a*b = 0, what do we know about a or b? To contract, if a*b = 4, what do we NOT know about either a or b?

    • one year ago
  8. marasofia1616 Group Title
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    what each of the equal?

    • one year ago
  9. marasofia1616 Group Title
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    them*

    • one year ago
  10. tkhunny Group Title
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    Kind of. a*b = 0 tells us that either a = 0 or b = 0. What can we tell about a and b when it isn't zero? a*b = 4, for example? Is one of them zero? Aboslutely not, but that's about all we can tell. Is a = 4? Maybe. Is b = 12? Maybe. Not a clue. Only "=0" is particularly helpful. So, what should we do with that factored expression?

    • one year ago
  11. chaguanas Group Title
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    tkhunny your explanations are more complex than the problem itself :)

    • one year ago
  12. tkhunny Group Title
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    That's why we invent notation, so that problems can be simpler than explanations. If we have this: Factor1 * Factor2 = 0 Then we must have Either this Factor1 = 0 or this Factor2 = 0 How can we apply this to your factored equation?

    • one year ago
  13. marasofia1616 Group Title
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    cos x+1(0)=0 I am not sure

    • one year ago
  14. tkhunny Group Title
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    \((\cos(x)+1)^{2} = 0\) We have two factors, but hey are exactly the same. This limits the possible results. \(\cos(x) + 1 = 0\) \(\cos(x) = -1\) \(x = \dfrac{3\pi}{2} + 2k\pi\) where \(k\in \mathbb{Z}\) Do you know what all that means?

    • one year ago
  15. marasofia1616 Group Title
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    I do... but if cos(x)=-1 wouldn't [x=\pi\]

    • one year ago
  16. marasofia1616 Group Title
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    \[x=\]

    • one year ago
  17. marasofia1616 Group Title
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    x=pi*

    • one year ago
  18. marasofia1616 Group Title
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    anyway I got it ... thanks for the help!

    • one year ago
  19. tkhunny Group Title
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    Awesome. How did I do that?! Right you are. \(x = \pi + 2k\pi\)

    • one year ago
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