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.