marigirl
  • marigirl
e^-x cos x – e^-x sinx=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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marigirl
  • marigirl
\[e^x(\cos(x)-\sin(x))=0\]
marigirl
  • marigirl
how can i solve for x now Please help me :)
zepdrix
  • zepdrix
The exponential is negative power? like this?\[\large\rm e^{-x}(\cos x-\sin x)=0\]

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marigirl
  • marigirl
oh yes i missed writing the negative. how can i solve it now.
zepdrix
  • zepdrix
Apply your `Zero-Factor Property`: \(\large\rm (a)(b)=0\implies a=0~or~b=0\)
zepdrix
  • zepdrix
So,\[\large\rm e^{-x}=0 \qquad\qquad\qquad (\cos x-\sin x)=0\]
marigirl
  • marigirl
i am not sure how to do the cos (x)-sin(x) part
zepdrix
  • zepdrix
Hmmm, I would probably add sin x to the other side. and then ummm...\[\large\rm \cos x=\sin x\]No ideas from here? :)
marigirl
  • marigirl
cos (x)+sin(x)=1
zepdrix
  • zepdrix
No no, \(\large\rm \cos^2x+\sin^2x=1\) Not \(\large\rm \cos x+\sin x=1\) :D
zepdrix
  • zepdrix
Lot of ways to proceed from here. Maybe the most straight forward approach would be to divide by cosine.
zepdrix
  • zepdrix
\[\large\rm 1=\frac{\sin x}{\cos x}\]
zepdrix
  • zepdrix
Do you remember an identity for sine over cosine? :o
marigirl
  • marigirl
cot(x)
marigirl
  • marigirl
pi/4 is the answer! thanks!
zepdrix
  • zepdrix
Hmm cotangent is the one with co on top :d you need tangent. but ya.. k cool :3
zepdrix
  • zepdrix
You need to recall that the exponential function has an asymptote at zero. So this other part is actually never true: \(\large\rm e^{-x}=0\)
marigirl
  • marigirl
if i divide by sin(x) ..
marigirl
  • marigirl
oh yup

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