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starwars18

  • one year ago

Find all solutions in the interval [0, 2π). cos^2(x) + 2 cos x + 1 = 0

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  1. starwars18
    • one year ago
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    Could it be simplified to (cos(x) + 1)^2?

  2. starwars18
    • one year ago
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    and i am totally confused how to do it after that

  3. starwars18
    • one year ago
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    Answers are 2pi, pi, pi/4 and 7pi/4, or pi/2 and 3pi/2

  4. Vocaloid
    • one year ago
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    your first step is correct, the next step is to take the square root of both sides

  5. xapproachesinfinity
    • one year ago
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    look at it like \(y^2+2y+1=0\)

  6. starwars18
    • one year ago
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    so cos(x) = -1 ???

  7. xapproachesinfinity
    • one year ago
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    solve for y tell you found the values then replace y by cosx then solve

  8. xapproachesinfinity
    • one year ago
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    oh yes

  9. Vocaloid
    • one year ago
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    yes! now find every x for which cos(x) = -1

  10. Pawanyadav
    • one year ago
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    Solve by putting y=cosx At last check values of y only in range of cosx then you will get the answer.

  11. starwars18
    • one year ago
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    pi?

  12. Vocaloid
    • one year ago
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    yup, good job!

  13. starwars18
    • one year ago
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    sorry my computer froze for a bit

  14. starwars18
    • one year ago
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    ty for your help

  15. xapproachesinfinity
    • one year ago
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    what if we extended the domain to (0, 5pi) what are the solutions ?

  16. xapproachesinfinity
    • one year ago
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    @starwars18

  17. starwars18
    • one year ago
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    -1 + n2pi, n being a real integer from 0 to 2

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