Solve equation for all solutions $\tan(3x)+1=0$ I am working on proving the $\tan(3x) = -1 $ for all solutions.  The big thing in the problem that is new to me is the substitution that has to be done. After the substitution I am not sure how the statement $\tan(3x) = -1$ is true. I am not sure what $3x$ means?  Also why is the $\tan(3x) = -1$ and not 1?  I feel like the negative sign is dropped and would like to find out more about how the that happens.  I also have a question about how I multiplied both sides by $\frac13$. Here are the steps I have for solving the problem.
$\tan(3x) = -1$
Let the $\tan(3x) = \tan(\theta)$.   Here is where the substitution begins.
The next steps are to express all solutions of $\tan(\theta) = -1 $ in notation form and to begin to finish the substitution back to the $\tan(3x)=-1$;
$ x = \frac{3 \pi}{4} + \pi k; \phantom1 k \in \mathbb{Z}$
$ 3x = \frac{3 \pi}{4} + \pi k; \phantom1 k \in \mathbb{Z}$
$ x = \frac{\pi}{4} + \pi k; \phantom1 k \in \mathbb{Z}$  I think this is the answer.
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*Here is where I think the negative sign is dropped
I was wondering how the substitution works when the $\tan(\frac{\pi}{4}) = 1 $ and not -1.  How does the $\tan(3x) = -1$?  What is the $\tan(3x)$ mean?
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Also now that I think of it when I solve for x by multiplying both sides by $\frac13$ would the equation be $ x = \frac{\pi}{4}+ \frac{\pi k}{3}; \phantom1 k \in \mathbb{Z}$?  By doing this however, it seems to defeat the domain of tangent.  The domain of tangent is $\pi$.  Would this be why the $\pi k $ is not included in the algebraic rules of multiplying both sides of the equation by $\frac13$.  Do I have a mathematical error too?
I have a second answer that I was wondering about.  The $\tan(\theta) = -1$ can also be $\frac{7 \pi}{4}$.  I was wondering why that answer gets dropped?  My thought is the $\frac{7 \pi}{4}$ has already been expressed in the equation $x = \frac{3 \pi}{4} + \pi k; \phantom1 k \in \mathbb{Z}$.
If k = 1 in the equation above, then $x = \frac{7 \pi}{4}$.  Writting this out long hand then doesn't need to be done for all solution because in doing so I would only be repeating $x = \frac{3 \pi}{4} + \pi k; \phantom1 k \in \mathbb{Z}$.
I have my doubts on the rational that one answer is dropped because it is already expressed.  I did complete the substitution for $\frac{7 \pi}{4}$ and after I did I found out the answer wasn't part of the solution.  I think it has something to do with what is meant by the $\tan(3x) = -1$.  I am not sure what the 3x is.  But here is the solution that isn't part of the answer and a few steps leading up to it;
$ x = \frac{7 \pi}{4} +\pi k; \phantom1 k \in \mathbb{Z}$, this is true for the $\tan(\theta) = -1$
$3x = \frac{7 \pi}{4} + \pi k; \phantom1 k \in \mathbb{Z} $  Begin to substitute back in $tan(3x) = -1$
$x = \frac{7 \pi}{12} + \pi k; \phantom1 k \in \mathbb{Z}$
This equation ends up not to be part of the solution.  I was wondering why?  Is it because it only repeats what has been said or is it some other reason?
 A: To solve an equation of the form $\tan(Ax)=b$ one correct form of the answer is always
$$ x=\frac{\arctan(b)}{A}+\frac{\pi}{A}k$$
Now it is true that $-\frac{\pi}{2}<\arctan(b)<\frac{\pi}{2}$ but you can use any other angle besides $\arctan(b)$ so long as its tangent is $b$. If you add any integer multiple of $\pi$ to $\arctan(b)$ that will also be an angle whose tangent is $b$.
Notice that when you divide by $A$ you must divide both terms by $A$, not just the first term.
Although the period of $\tan(x)$ is $\pi$, the period of $\tan(Ax)$ is $\frac{\pi}{A}$.
A: Okay.

I am not sure what 3x means?

It means $3$ times $x$.

Also why is the tan(3x)=−1 and not 1?

$x$ is the value so that AFTER you multiply $x$ by $3$ to get $3x$, and then you get the result of $\tan(3x)$ that result will be $-1$.   It does not matter what the result of $\tan x$ is (which will be $+1$ one-third of the time).
This would be similar to $5x + 4 = 19$ so $x=3$ but then asking: Since $3 + 4$ why isn't it $5x+4 = 7$ as $x+4 = 7$.  The reason is that $5x$ is a different thing than $x$.

Also now that I think of it when I solve for x by multiplying both sides by $\frac 13$ would the equation be $x=\frac π4+\frac{πk}3;k∈Z$?

Yes, that is correct.

By doing this however, it seems to defeat the domain of tangent.

I think you mean it defeats to period of tangent.  The period of tangent is $\pi$ so that if $\tan \theta = w$ then $\tan (\theta + k\pi) = \theta$.  But the function we are looking at is not $f(x) = \tan x$.  The function we are looking at is $f(x) = \tan (3x)$.  That is a different function and it has different period. (And it would have a different domain if the domain of $\tan$ where restricted in any way).
So can see that the period of $f(x) = \tan (3x)$ is $\frac \pi 3$ because if $\tan (3x) = w$ then $\tan (3(x + \frac \pi 3)) = \tan (3x + 3\cdot \frac \pi 3) = \tan (3x + \pi) = \tan (3x)$.
...
So to solve your problem.
Let $\theta = 3x$.
$\tan 3x = -1$
$\tan \theta = -1$ so
$\theta = \frac {3\pi} 4 + k\pi$
$3x = \frac {3\pi} 4 + k\pi$
$x =\frac {\pi} 4 + k\frac \pi 3$.
And we can verify that is correct:
If $x = \frac{\pi}4 + k\frac \pi 3$ then
$\tan 3x = \tan (3(\frac \pi 4 + k\frac \pi 3))=$
$\tan( \frac {3\pi}4 + k\pi)= -1$.
It does not matter that if $x = \frac \pi 4$ than $\tan x = 1$ or that if $x = \frac \pi 4 +\frac \pi 3$ then $\tan x = -3.73.....$ because what $\tan x$ might actually be was never part of the problem.  All we care about is what happens AFTER we multiply $x$ by $3$.

The tan(θ)=−1 can also be $\frac{7π}4$. I was wondering why that answer gets dropped? My thought is the $\frac{7π}4$ has already been expressed in the equation $x=\frac {3π}4+πk;k∈Z$.

That is correct.  $\frac {7\pi}4 = \frac {3\pi}4 + \pi$.
But be careful.  You NEVER had the equation $x = \frac {3\pi}4 + k\pi$.  You had the equation that $\huge{3}$$x = \frac {3\pi}4 + k\pi$.

I did complete the substitution for $\frac{7π}4 and after I did I found out the answer wasn't part of the solution

That is because $3\cdot \frac {7\pi}4 = \frac {21\pi}4$ and $\tan \frac {21\pi}4\ne -1$.
Again we are not concerned if $\tan something = -1$.  We are concerned if $\tan 3\times something = -$ and $\tan 3\times \frac {7\pi}4 \ne -1$.

$x=\frac{7π}4+πk;k∈Z$, this is true for the tan(θ)=−1

Not $x$.... $\theta$.
$\theta =\frac {7\pi}4 + \pi k$.

$3x=\frac{7π}4+πk;k∈Z$
Begin to substitute back in tan(3x)=−1

This is correct.

$x=\frac {7π}{12}+πk;k∈Z$

No, you need to distribute the $\frac 13$
$x = \frac {7\pi}{12} + k\frac \pi 3$.
And that works out fine as $\frac {7\pi}{12} =\frac {\pi}4 + \frac {\pi}3$
A: For any $h, k \in \mathbb{R}$, the general solution of the equation
$$\tan (h x)=k \quad$$ is
$$
\begin{aligned}
h x &=n \pi+\tan ^{-1} k \\
x &=\frac{n \pi}{h}+\frac{\tan ^{-1} k}{h}
\end{aligned}
$$
where $\tan ^{-1} k \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right).$
Noting that
$$
\tan ^{-1}(-1)=-\frac{\pi}{4},
$$
we have
$$
\begin{aligned}
\tan 3 x+1 &=0 \\
\tan 3 x &=-1 \\
3 x &=n \pi-\frac{\pi}{4}, \text { where } n \in Z. \\
x &=\frac{4 n-1}{12} \pi
\end{aligned}
$$
which is the general solution of the equation.
Wish it helps.
