Question

e^-x cos x – e^-x sinx=0

Answer 1

\[e^x(\cos(x)-\sin(x))=0\]

Answer 2

how can i solve for x now
Please help me :)

Answer 3

The exponential is negative power? like this?\[\large\rm e^{-x}(\cos x-\sin x)=0\]

Answer 4

oh yes i missed writing the negative. how can i solve it now.

Answer 5

Apply your `Zero-Factor Property`: \(\large\rm (a)(b)=0\implies a=0~or~b=0\)

Answer 6

So,\[\large\rm e^{-x}=0 \qquad\qquad\qquad (\cos x-\sin x)=0\]

Answer 7

i am not sure how to do the cos (x)-sin(x) part

Answer 8

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

Answer 9

cos (x)+sin(x)=1

Answer 10

No no, \(\large\rm \cos^2x+\sin^2x=1\)
Not \(\large\rm \cos x+\sin x=1\) :D

Answer 11

Lot of ways to proceed from here.
Maybe the most straight forward approach would be to divide by cosine.

Answer 12

\[\large\rm 1=\frac{\sin x}{\cos x}\]

Answer 13

Do you remember an identity for sine over cosine? :o

Answer 14

cot(x)

Answer 15

pi/4 is the answer! thanks!

Answer 16

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

Answer 17

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

Answer 18

if i divide by sin(x) ..

Answer 19

oh yup