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?
Best Answer
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 $\frac{\pi}{A}$.