# Find the domain of a tangent function

Consider function $$f\left(x\right)\:=\:\tan\left(3x-\pi \right)$$. I know that $$\tan$$ is undefined at $$\frac{\pi}{2}+\pi k,\:k\in\mathbb{Z}$$. I solved $$3x-\pi=\frac{\pi}{2}$$, which means that $$x$$ must never be $$\frac{\pi}{2}$$.

However, the proposed solution for this problem is $$\left\{x\in \mathbb{R}:x\:\text{is different from}\:\frac{\pi }{6}+\frac{k\pi }{3},\:k\in \mathbb{Z}\right\}$$. I can't quite understand this conclusion.

• First of all just see that $\tan(3x-\pi)=-\tan(\pi-3x)=\tan 3x$. Nov 1, 2018 at 16:41

$$\tan(3x-\pi)$$

Recall the period of $$\tan \theta$$ is $$\pi$$.

$$\tan(3x-\pi) = \tan(3x)$$

Now, finding the domain is simple. For all $$n \in \mathbb{Z}$$, the following is concluded.

$$3x \neq \frac{\pi}{2}+\pi n$$

$$x \neq \frac{\pi}{6}+\frac{\pi n}{3}$$

$$3x-\pi \neq \frac{\pi}{2}+\pi n \implies 3x \neq \frac{\pi}{2}+(n+1)\pi$$
$$x \neq \frac{\pi}{6}+\frac{(n+1)\pi}{3}$$
$$n$$ covers all integer values. So does $$n+1$$, so your domain is essentially the same.
You are correct that $$3x-\pi\neq \frac\pi 2$$. However, the equation you should solve is $$3x-\pi = \frac\pi 2 + \pi k$$ for $$k\in\Bbb Z$$. This becomes $$3x=\frac{3\pi}2+\pi(k+1)$$ and so $$x=\frac\pi 2+\frac\pi 3(k+1)$$. Now, suppose $$n-2=k$$. As $$k$$ is just some arbitrary integer, $$n$$ is also just an arbitrary integer so we still have all the same solutions. Substitutin gives that $$x=\frac\pi 2 +\frac\pi 3(n-2+1)=\frac\pi 2 + \frac\pi 3(n-1)= \frac\pi 2 -\frac\pi 3 + \frac\pi 3n = \frac\pi 6+\frac\pi 3 n$$ for $$n\in\Bbb Z$$.