Is the Trigonometric Equation {3}^{tg2x}*{3}^{ctg3x}=0 Solvable?

[Physicsissuef]

Homework Statement

{3}^{tg2x}*{3}^{ctg3x}=0

Homework Equations

The Attempt at a Solution

Is this possible to solve?

I don't think so.

{3}^{tg2x+ctg3x}=0

For whatever value of x, we can't get 0, right?

 

[HallsofIvy]

you are correct. No value of x will make 3^x= 0 so that equation has no solution.

 

[Physicsissuef]

Ok, thank you.

 

[sutupidmath]

you are correct. No value of x will make 3^x= 0 so that equation has no solution.

If we suppose that x is from the extended reals x\in[-\infty,+\infty],

Could we say then that equations like a^{x}=0 have solution for x=-\infty, \ \ \ a\in R ??

 

[Tedjn]

It is true that \lim_{x\to-\infty}a^x = 0 when |a| > 1, but that is not the same as saying that x = -\infty.

 

[sutupidmath]

It is true that \lim_{x\to-\infty}a^x = 0 when |a| > 1, but that is not the same as saying that x = -\infty.

Yeah, i do understand this part, i was just wondering how does one perform opertations on the extended reals, that is [-\infty,+\infty], rather than just in (-\infty,+\infty)
.

 

[HallsofIvy]

I'm not aware of any way of way of definining trig functions on the extended reals.