e^-x cos x – e^-x sinx=0

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

e^-x cos x – e^-x sinx=0

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

\[e^x(\cos(x)-\sin(x))=0\]
how can i solve for x now Please help me :)
The exponential is negative power? like this?\[\large\rm e^{-x}(\cos x-\sin x)=0\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question