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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 $2$.

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

2 Answers 2

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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 $k\neq 4h+2$

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    $\begingroup$ For trig functions, use a backslash, eg $\sin x$ comes out as $\sin x$. $\endgroup$ Commented Mar 11, 2021 at 19:28
  • $\begingroup$ Excuse me for my delay...thank you very much. $\endgroup$
    – Sebastiano
    Commented Mar 13, 2021 at 13:00
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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 $k=(4h+1)\neq(4h+2)$

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

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