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marigirl

  • one year ago

e^-x cos x – e^-x sinx=0

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  1. marigirl
    • one year ago
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    \[e^x(\cos(x)-\sin(x))=0\]

  2. marigirl
    • one year ago
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    how can i solve for x now Please help me :)

  3. zepdrix
    • one year ago
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    The exponential is negative power? like this?\[\large\rm e^{-x}(\cos x-\sin x)=0\]

  4. marigirl
    • one year ago
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    oh yes i missed writing the negative. how can i solve it now.

  5. zepdrix
    • one year ago
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    Apply your `Zero-Factor Property`: \(\large\rm (a)(b)=0\implies a=0~or~b=0\)

  6. zepdrix
    • one year ago
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    So,\[\large\rm e^{-x}=0 \qquad\qquad\qquad (\cos x-\sin x)=0\]

  7. marigirl
    • one year ago
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    i am not sure how to do the cos (x)-sin(x) part

  8. zepdrix
    • one year ago
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    Hmmm, I would probably add sin x to the other side. and then ummm...\[\large\rm \cos x=\sin x\]No ideas from here? :)

  9. marigirl
    • one year ago
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    cos (x)+sin(x)=1

  10. zepdrix
    • one year ago
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    No no, \(\large\rm \cos^2x+\sin^2x=1\) Not \(\large\rm \cos x+\sin x=1\) :D

  11. zepdrix
    • one year ago
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    Lot of ways to proceed from here. Maybe the most straight forward approach would be to divide by cosine.

  12. zepdrix
    • one year ago
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    \[\large\rm 1=\frac{\sin x}{\cos x}\]

  13. zepdrix
    • one year ago
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    Do you remember an identity for sine over cosine? :o

  14. marigirl
    • one year ago
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    cot(x)

  15. marigirl
    • one year ago
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    pi/4 is the answer! thanks!

  16. zepdrix
    • one year ago
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    Hmm cotangent is the one with co on top :d you need tangent. but ya.. k cool :3

  17. zepdrix
    • one year ago
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    You need to recall that the exponential function has an asymptote at zero. So this other part is actually never true: \(\large\rm e^{-x}=0\)

  18. marigirl
    • one year ago
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    if i divide by sin(x) ..

  19. marigirl
    • one year ago
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    oh yup

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