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

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 $$2$$.

• We can write : $tan(\pi+6x)=tan(2k\pi-2x)\Rightarrow x=(2k-1)\pi+\frac{\pi}{8}$ Commented Mar 11, 2021 at 17:31
• @Sebastinao, please see my answer (with the constrain k≠ 4h+2). Commented Mar 11, 2021 at 19:10
• @sirous Hi, yes you have right. But how I found $k≠ 4h+2$? Commented Mar 11, 2021 at 19:38
• @BrightStar Yes I have seen but there are many mistakes :-) in $\LaTeX$ :-( +1 for your answer. Commented Mar 11, 2021 at 19:39
• It is probably $k\neq(4h+2)$ for general solution $x=k\pi+\frac{\pi}8$. Commented Mar 12, 2021 at 8:26

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 $$k\neq 4h+2$$

• For trig functions, use a backslash, eg $\sin x$ comes out as $\sin x$. Commented Mar 11, 2021 at 19:28
• Excuse me for my delay...thank you very much. Commented Mar 13, 2021 at 13:00

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

• Very nice also your answer. +1 Commented Mar 12, 2021 at 11:45
• Hi, I have not understood the reason of this downvote on my question. Commented Mar 13, 2021 at 10:34