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mgarcia634

  • 3 years ago

Find all solutions in the interval [0, 2π). cos2x + 2 cos x + 1 = 0

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  1. mgarcia634
    • 3 years ago
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    cos^2 x+ 2 cos x +1=0

  2. cwrw238
    • 3 years ago
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    this is a quadratic equation which will factorise

  3. Stiwan
    • 3 years ago
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    \[\cos²(x) + 2\cos(x) + 1 = (\cos(x) + 1)²\] Hence the roots are all x out of [0,2pi) for which cos(x) = -1. -pi and pi

  4. Stiwan
    • 3 years ago
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    Sorry, of course it's just + pi as there are no negative numbers in this interval

  5. cwrw238
    • 3 years ago
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    what about 0 as a solution?

  6. cwrw238
    • 3 years ago
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    no sorry - its -1

  7. cwrw238
    • 3 years ago
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    cos 0 = 1 of course

  8. mgarcia634
    • 3 years ago
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    the answer choices are a) x=2pi B)x=pi c) x=pi/4, 7pi/4

  9. cwrw238
    • 3 years ago
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    are you sure you have the correct question is it cos 2x or cos^2 x ?

  10. mgarcia634
    • 3 years ago
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    it is cos^2 x

  11. Stiwan
    • 3 years ago
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    It's B) x = pi as in my answer.

  12. cwrw238
    • 3 years ago
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    in that case the answer is as Stiwam said

  13. mgarcia634
    • 3 years ago
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    thankyou!

  14. cwrw238
    • 3 years ago
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    yw

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