Basic Calculus: Solving for x in sinx(lnx)=0 over [0, 2pi]

[name_ask17]

Homework Statement

If sinx(lnx)=0 over [0, 2pi] then x=_____

The Attempt at a Solution

My attempt was to move sin over to the other side to get lnx= 0/sinx and then get lnx=0, making x=1. But is it incorrect to just move sinx over to the other side, because if i move lnx over to the other side, i would get x=0. Can you please explain to me what I am doing wrong here?

 

[Hells]

Well, if those are factors; Sin(x) and ln(x). Then there are 4 solutions. You found ln(1)=0 then you just need to find those values that makes Sin(x)=0, one of the solutions are not legit since they are not defined by ln(x)

The question has nothing to do with calculus though :P

 

[name_ask17]

how do you know there are 4 solutions?
and yes, this is just our summer review to go into calc. lol

 

[eumyang]

Then you should have posted this in the Precalculus subforum instead.

There are 3 solutions because,
there is 1 solution in [0, 2π] where ln x = 0, and
there are 2 solutions in [0, 2π] where sin x = 0 (a 3rd is not in the domain of the ln x function).

 

[Mark44]

Then you should have posted this in the Precalculus subforum instead.

There are 4 solutions because,
there is 1 solution in [0, 2π] where ln x = 0, and
there are 3 solutions in [0, 2π] where sin x = 0.

I get 3 solutions. One of the 3 solutions in [0, 2π] for which sin x = 0 is not in the domain of the ln function.

 

[QuarkCharmer]

You can't really just move the sine over like that. Sin is an operator! What you need to do is consider what would make sin(something) = 0. (consult your unit circle). Once you have those 2 solutions, set the argument of the sin to that.
ln(x) = solution 1
ln(x) = solution 2
etc.

 

[Mark44]

If sinx(lnx)=0 over [0, 2pi] then x=_____

You can't really just move the sine over like that. Sin is an operator! What you need to do is consider what would make sin(something) = 0. (consult your unit circle). Once you have those 2 solutions, set the argument of the sin to that.
ln(x) = solution 1
ln(x) = solution 2
etc.

QuarkCharmer, you're misreading the problem, which is understandable due to the way the OP wrote the problem. Howebver, the expression on the left is a product, not a composition. It's sin(x) * ln(x), not sin(ln(x)).

 

[QuarkCharmer]

My mistake, feel free to delete my post(s)

 

[eumyang]

One of the 3 solutions in [0, 2π] for which sin x = 0 is not in the domain of the ln function.

Ack, that's what I get for blinding following someone else's post. Post corrected.