Condensing of $f(x)=\sec(\frac{\pi}{4}x)$ The function $f(x)=\sec(\frac{\pi}{4}x)$ is undefined every time $\cos(\frac{\pi}{4}x)$is equal to $0$. Therefore, I set cosine equal to zero, and the solutions I got were: > $x=2+8k$ and $x=6+8k$ $$k\in\mathbb{Z}$$ I decided to condense the solutions into one, and got: > $$x=2+4k$$ $$k\in\mathbb{Z}$$ Is the way I condensed it correct? If it is, is this how you would define where $\sec(\frac{\pi}{4}x)$ is continuous? > $$x\in\mathbb{R}\lvert x\ne2+4k\lvert k\in\mathbb{Z}$$

To rigorously show that condensation, you need to show that $\\{8k+2\\}\cup\\{8k+6\\} =\\{4k+2\\} $.

Here's how I would do that seemingly obvious proof:

If $n = 8k+2$, then $n = 4(2k)+2$, so $n \in \\{4k+2\\} $, so $\\{8k+2\\} \subset \\{4k+2\\} $.

Similarly, if $n = 8k+6$, then $n = 4(2k)+4+2 = 4(2k+1)+2 $, so $n \in \\{4k+2\\} $, so $\\{8k+6\\} \subset \\{4k+2\\} $.

Therefore $\\{8k+2\\}\cup\\{8k+6\\} \subset \\{4k+2\\} $.

To go the other way, if $n = 4k+2$, then $k$ is either even or odd, so either $k = 2j$ or $k = 2j+1$ for some $j$.

In the first case, $n = 4k+2 =4(2j)+2 =8j+2 $ so $n \in \\{8k+2\\} $.

In the second case, $n = 4k+2 =4(2j+1)+2 =8j+6 $ so $n \in \\{8k+6\\} $.

Therefore $\\{4k+2\\} \subset \\{8k+2\\}\cup\\{8k+6\\} $.

As to where $\sec(\frac{\pi}{4}x)$ is continuous, be sure that you include negative $k$.