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TNNG

  • one year ago

Find all solutions to the equation. cos x = sin x

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  1. T0mmy
    • one year ago
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    Doesn't make sense

  2. mathmate
    • one year ago
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    When in doubt, draw a sketch: |dw:1357770541986:dw| The sketch is for 0<=x <=2pi. The general solution requires the form x=.... +2k pi where k is an integer.

  3. TNNG
    • one year ago
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    still do not get it

  4. mathmate
    • one year ago
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    You will need to solve for sin(x)=cos(x) between 0 and 2pi. The same solution will be valid if you add 2k pi to the obtained solution. Are you able to find the solution between 0 and 2pi?

  5. TNNG
    • one year ago
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    look is the answer pi/2 +n(pi)(0,+/-1,+/- 2

  6. BluFoot
    • one year ago
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    Looking at the unit circle is another way to do it if you've learned that. When are the cos and sin values identical?

  7. TNNG
    • one year ago
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    I tried about everything what is the answer

  8. cinar
    • one year ago
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    |dw:1357780533732:dw|

  9. TNNG
    • one year ago
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    so it is pi/2 + n(pi)

  10. cinar
    • one year ago
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    http://www.wolframalpha.com/input/?i=solve+cosx-sinx%3D0

  11. cinar
    • one year ago
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    yes one solution is that..

  12. TNNG
    • one year ago
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    ok thank you i think i get it now

  13. cinar
    • one year ago
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    I meant pi/4+2npi

  14. TNNG
    • one year ago
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    the other is 7pi /4 + 2npi

  15. BluFoot
    • one year ago
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    it should be pi/4 +2npi and 5pi/4 + 2npi

  16. TNNG
    • one year ago
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    i dont have a answer like that it is just pi/4 +2npi and 7pi /4 + 2npi. the answer with 5pi/4 + 2npi is paired with 3pi/4 + 2npi

  17. BluFoot
    • one year ago
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    I don't know what the question is exactly but the solution to sinx=cosx is definitely x=pi/4 +2npi and x=5pi/4 + 2npi

  18. TNNG
    • one year ago
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    Find all solutions to the equation. cos x = sin x is the question they are asking

  19. BluFoot
    • one year ago
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    Well then their answer is wrong!

  20. mathmate
    • one year ago
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    |dw:1357774360119:dw|

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