Solve Trig Equation: tan^2x-3=0 | Homework Help

[hominid]

Homework Statement

Solve for x

Homework Equations

$$tan^2x-3=0$$

The Attempt at a Solution

$$tan^2x-3=0$$

$$tanx=\pm\sqrt{3}$$

I'm not sure what to do after this. I could the $$tan^{-1}(\sqrt{3})=x$$ or $$x=\frac{\pi}{3}$$

But then what do I do with the $$-\sqrt{3}$$?

 

[Mentallic]

So $tan^2x=\pm \sqrt{3}$

This means you have to solve 2 equations, mainly:
$tanx=\sqrt{3}$ and $tanx=-\sqrt{3}$

Let's place a restriction on the possible values of x since the tangent function is periodic (it repeats itself every $\pi$)

So we'll put the domain as $0\leq x\leq 2\pi$

Now, you've already found the first value, $x=\pi /3$
but there is another solution in the domain. Do you know about that ASTC or whatever?

For the second equation, you'll still use the same triangle with opposite:adjacent side $\sqrt{3}:1$, so the $\pi /3$ will still be in play, and you just have to use the same idea that the ASTC thing represents. If you don't know what that is, just ask and we'll explain it.

 

[hominid]

Wonderful. So you leave out $tanx=-\sqrt{3}$ because of $0\leq x\leq 2\pi$ but wait, you said it has to be $0\leq x\leq \pi$ because it's a tangent function.


Then you're left with $x=\frac{\pi}{3}+{\pi}n$ & $x=\frac{2\pi}{3}+{\pi}n$

Is this right?

Thanks for the help

 

[Mentallic]

No, I never said anything like that. It seems like you have a lot of concepts mixed up. It's also puzzling why you know how to express a trigonometric function's infinite periodicity, but you've failed to show an understanding of the basics.
Anyway...

I would recommend you study the basics of trigonometry again (no offense intended). Or, if you can't be bothered with that, I think we can get a try clear up the fuzzy picture in your head by getting you to stare at the tangent function for a good minute

So you leave out $tanx=-\sqrt{3}$ because of $0\leq x\leq 2\pi$

For any value $0\leq x\leq 2\pi$ can $tanx<0$ ? If so, which values? More specifically, where would $tanx=-\sqrt{3}$ ?

but wait, you said it has to be $0\leq x\leq \pi$ because it's a tangent function.

I said that the tangent function is periodic every $\pi$.
i.e. if $f(x)=tanx$ then $f(x+\pi n)=f(x)$ for any integer n

Then you're left with $x=\frac{\pi}{3}+{\pi}n$ & $x=\frac{2\pi}{3}+{\pi}n$

The first is correct - although the restriction on x I placed just for ease is void now, but it's fine. Where did the second result come from?
I'm going to assume you don't know what the ASTC thing is since you ignored it completely. For $0\leq x\leq \pi$ all values of $tanx$ are unique, and remember the tan function is period about $\pi$, so what does this say about your answer to $tanx=\sqrt{3}\Rightarrow x=\pi/3,2\pi/3$ ?

If the first solution to $tanx=\sqrt{3}$ is $x=\pi/3$ and tanx is periodic about $\pi$, then what is the next solution?
Just use the general solution you've provided: $x=\pi/3+\pi n$

 

[hominid]

I'm trying to refresh my memory since my high school trig class ten years ago. I understand what ASTC is, what the function looks like, etc. With all due respect, I think you're overlooking what I did.

I don't see why you're restricting the domain to $2\pi$ when a tangent function has a period of $\pi$. And since that is the domain, we don't want negative values, thus $tanx=-\sqrt{3}=-\frac{\pi}{3}$ doesn't work, but if you add $\pi$ to that value you get $\frac{2\pi}{3}$.

And since trig functions are periodic, and the tan has a period of pi, I add 2pi to each of the solution and multiply by any integer n.

You're making it sounds like I made a mistake, and as far as I can tell, you didn't say that was wrong, or show that it was incorrect.

If I am missing something, I appreciate the help.