Procedure or technique or reasoning to get $2+4h$?

Question

I have solved for my students of an high school this simple trigonometric equation:

$$\tan(\pi+6x)=-\tan(2x)\tag 1$$

The $(1)$ is equivalent to (I remember also that $\tan(\alpha)=\tan(\mathbb Z\pi+\alpha)$)

$$\tan(\pi+6x)=\tan(-2x)\tag 2 \iff x=\frac{\pi}{8}k^*, \quad k^*=k-1\in\Bbb Z$$

But the solution of the textbook is $x=k\pi/8$ with $k\neq 4h+2$.

How can I find the value $\color{red}{4h+2}$?

If I calculate the domain I will have

$$\begin{cases} x \neq -\dfrac \pi{12}+\Bbb Z\dfrac \pi{6}\\[0.5em] \tag 3 x\neq \dfrac \pi4+\Bbb Z\dfrac\pi2 \end{cases}$$

I have done the tests and the condition $4h+2$ is equivalent to $(3)$. Just a curiosity looking the 2nd negation of the $(3)$ I have the denominator $4$ and $2$.

Answer 1

The constrains come from the domain of tangent function. The domains of $\tan x$ excludes $k\pi+\frac{\pi}{2}$, there are constrains placed on $\tan2x$ and $\tan 6x$, where $k,m,n,h \in \mathbb Z$.

For $\tan 2x$, we have $$2x \neq m\pi +\frac{\pi}{2}$$ with $x\neq \frac{(2m+1)\pi}{4}$.

For $\tan 6x$, we have $$6x \neq n\pi+\frac{\pi}{2}$$ with $x \neq \frac{(2n+1)\pi}{12}$.

Or

$k\pi/8 \neq\frac{(2m+1)\pi}{4}$, $k\neq 2(2m+1) =4m+2, \quad (A)$;

$k\pi/8 \neq\frac{(2n+1)\pi}{12}$, $k\neq \frac{2(2m+1)}{3} \equiv \frac{4m+2}{3} \quad (B)$.

If (A) is satisfied, so is (B).

Therefore, The solutions are

$$x = \frac{k\pi}{8}$$ with $k\neq 4h+2$

Answer 2

Another approach:

For $x=\pi+\frac{\pi}8$:

LHS: $\tan [\pi +6(\pi+\frac{\pi}8)]=\tan(\pi+\frac{3\pi}4)=\tan(-\frac{\pi}4)$

RHS: $-2(\pi+\frac{\pi}8)=- \frac{\pi}4$

So general form of solution can be:

$x=(2m+1)\pi+\frac{\pi}8=[8(2m+1)+1]\frac{\pi}8$

Let $2(2m+1)=h$, we have:

$x=(4h+1)\frac{\pi}8$

So $k=(4h+1)\neq(4h+2)$