## marigirl one year ago e^-x cos x – e^-x sinx=0

1. marigirl

$e^x(\cos(x)-\sin(x))=0$

2. marigirl

3. zepdrix

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

4. marigirl

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

5. zepdrix

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

6. zepdrix

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

7. marigirl

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

8. 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? :)

9. marigirl

cos (x)+sin(x)=1

10. zepdrix

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

11. zepdrix

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

12. zepdrix

$\large\rm 1=\frac{\sin x}{\cos x}$

13. zepdrix

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

14. marigirl

cot(x)

15. marigirl

16. zepdrix

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

17. zepdrix

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

18. marigirl

if i divide by sin(x) ..

19. marigirl

oh yup