Help on solving this equation!

Dear all,

I am having trouble solving this equation, it has been some time since I have done this kind of stuff.

Is there anyone here that can tell me how to solve this equation? It is very basic, but it has been some time since I have done these equations.

Cos(3*pi*x)>ln(x+1)

I just can not get my head around it, I only came this far (Headbang):

e^{cos(3*pi*x)}>x+1

Also, I am having trouble trying to get the integral of sin(x)e^{cos(x)}. How does this work with either the reverted product rule or the substitution method?

Thank you!

Re: Help on solving this equation!

Quote:

Originally Posted by

**Maxime** Dear all,

I am having trouble solving this equation, it has been some time since I have done this kind of stuff.

Is there anyone here that can tell me how to solve this equation? It is very basic, but it has been some time since I have done these equations.

Cos(3*pi*x)>ln(x+1)

I just can not get my head around it, I only came this far (Headbang):

e^{cos(3*pi*x)}>x+1

Also, I am having trouble trying to get the integral of sin(x)e^{cos(x)}. How does this work with either the reverted product rule or the substitution method?

Thank you!

For the integral, write $\displaystyle \displaystyle \begin{align*} \int{\sin{(x)}e^{\cos{(x)}}\,dx} &= -\int{-\sin{(x)}e^{\cos{(x)}}\,dx} \end{align*}$ and make the substitution $\displaystyle \displaystyle \begin{align*} u = \cos{(x)} \implies du = -\sin{(x)}\,dx \end{align*}$ and the integral becomes $\displaystyle \displaystyle \begin{align*} -\int{e^u\,du} \end{align*}$. Go from here.

Re: Help on solving this equation!

Thank you! That example really brings out memories of the substitution method.

Anyone any idea of how to solve this for x?:

cos(3*pi*x)>ln(x+1)

Re: Help on solving this equation!

You're not going to be able to get exact solutions. I'd suggest getting wolfram to graph both equations, noting where they cross, and deciding which intervals satisfy your inequality.