What is the solution to (sinx)*(lnx)= 0 over [0, 2pi]?

[name_ask17]

Homework Statement

If (sinx)*(lnx)= 0 over [0, 2pi], then what does x equal?

Ok, I know I posted a question similar to this one a few days back, but I need some clarification on something. Below is the work that someone showed me.

lnx=sin^-1 (0)
lnx=0
Log\hat{}x\check{}e=0 (If you don't understand that, it is log and the superscript is x, subscript is e)
x=1

First of all, is that correct? Does x=1?

And if so, how did they get the sin inverse over to the right hand side of the equation, because you are not dividing it by 1, you are dividing it by 0, which would just make it 0, not sin inverse. Please clarify if you can. Thanks in advance.

 

[gb7nash]

You're looking at this way too hard. You're multiplying two expressions together, so the only way you're going to get 0 is if sinx = 0 or lnx = 0.

For what value of x is sinx = 0 from [0, 2pi]? For what value of x is lnx = 0 from [0, 2pi]?

 

[SammyS]

Homework Statement

If (sinx)*(lnx)= 0 over [0, 2pi], then what does x equal?

Ok, I know I posted a question similar to this one a few days back, but I need some clarification on something. Below is the work that someone showed me.

lnx=sin^-1 (0) This is not correct !
...

First of all, is that correct? Does x=1?

And if so, how did they get the sin inverse over to the right hand side of the equation, because you are not dividing it by 1, you are dividing it by 0, which would just make it 0, not sin inverse. Please clarify if you can. Thanks in advance.

Well, x=1 is one solution, but you're right, that is not the way to use the arcsin function.

To solve (sin(x))*(ln(x))= 0 over [0, 2pi], use the zero product property.