# tan3x=tanx

$\begingroup$

I'm asked vĩ đại solve $\tan{x} = \tan{3x}$

Bạn đang xem: tan3x=tanx

Here's my attempt:

$$\tan{x} = \tan{3x}$$ $$\tan{x} = \tan{(x + 2x)}$$ $$\tan{x} = \frac{\tan{x} + \tan{2x}}{1-\tan{x}\tan{2x}}$$

Recall the identity: $$\tan{2x} = \frac{2\tan{x}}{1-\tan^2{x}}$$

So then we have: $$\tan{x} = \frac{\tan{x} + \frac{2\tan{x}}{1-\tan^2{x}}}{1-\tan{x}\frac{2\tan{x}}{1-\tan^2{x}}}$$ $$\tan{x} - \tan^2{(x)} \cdot \frac{2\tan{x}}{1-\tan^2{x}} = \tan{x} + \frac{2\tan{x}}{1-\tan^2{x}}$$ $$-\tan^2{(x)} \cdot \frac{2\tan{x}}{1-\tan^2{x}} = \frac{2\tan{x}}{1-\tan^2{x}}$$ $$-\tan^2{x} \cdot \frac{2\tan{x}}{1-\tan^2{x}} \cdot \frac{1-\tan^2{x}}{2\tan{x}} = 1$$ $$\tan^2{x} = -1$$

This does obviously not compute. Why is my way wrong and how can I go about solving it?

asked Aug 8, năm nhâm thìn at 22:00

$\endgroup$

1

$\begingroup$

Why vì thế you try vĩ đại make things more complex kêu ca they are? Formally,

\begin{align*}\tan x=\tan 3x&\iff 3x\equiv x\mod \pi \\ &\iff 2x\equiv0\mod \pi \iff x\equiv 0\mod\frac\pi2. \end{align*}

However, $\tan x$ is defined if and only if $x\not\equiv\dfrac\pi2\mod\pi$, ví the effective solutions are $$x\equiv 0\mod \pi.$$

answered Aug 8, năm nhâm thìn at 22:24

$\endgroup$

$\begingroup$

When you arrive at $$-\tan^2x \frac{2\tan x}{1-\tan^2x} = \frac{2\tan x}{1-\tan^2x}$$ you can't multiply both sides by $\frac{1-\tan^2x}{2\tan x}$, because this operation is allowed only when the multiplier is nonzero.

You should rather move everything vĩ đại the right-hand side and collect terms, getting $$0=\frac{2\tan x}{1-\tan^2x}(\tan^2x+1)$$ Since $\tan^2x+1\ne0$, you get $$\tan x=0$$ and ví $x=k\pi$.

You can also observe that $\tan3x=\tan x$ means $$3x=x+k\pi$$ so $$x=k\frac{\pi}{2}$$ provided $\tan x$ and $\tan3x$ exist, which is not the case when $k$ is odd. Thus the solutions are $x=(2h)\frac{\pi}{2}=h\pi$ ($h$ integer).

answered Aug 8, năm nhâm thìn at 22:28

$\endgroup$

4

$\begingroup$

You can't obviously cancel in the last (or one before the last, in fact) step, thus you must have

$$\frac{2\tan x}{1-\tan^2x}=0\iff \tan x=0\iff x=k\pi\;,\;\;k\in\Bbb Z.$$

answered Aug 8, năm nhâm thìn at 22:08

$\endgroup$

$\begingroup$

Your proof is fine up vĩ đại

$-\tan^2{(x)}(\frac{2\tan{x}}{1-\tan^2{x}}) = \frac{2\tan{x}}{1-\tan^2{x}}$

Then you vì thế $-\tan^2{x} \cdot (\frac{2\tan{x}}{1-\tan^2{x}}) \cdot (\frac{1-\tan^2{x}}{2\tan{x}}) = 1$.

Xem thêm: mg+hno3 đặc

But you can onlty vì thế that if $\frac{2\tan{x}}{1-\tan^2{x}} \ne 0$.

So you need vĩ đại say:

"Assume $\frac{2\tan{x}}{1-\tan^2{x}}\ne 0$ then

"$-\tan^2{x} \cdot (\frac{2\tan{x}}{1-\tan^2{x}}) \cdot (\frac{1-\tan^2{x}}{2\tan{x}}) = 1$

"$-\tan^2{x} = 1$.

"But this is impossible.

"So $\frac{2\tan{x}}{1-\tan^2{x}} = 0$"

And go on from there:

"So $\tan x = \sin x/\cos x = 0$.

"So $x = k\pi$".

=====

Or you can note $tan z = tan x \iff z = x + k\pi$.

So $3x = x + k\pi$ ví $2x = k\pi$ ví $x = \frac k 2 \pi$. But $\tan \frac k 2 \pi; k$ odd is undefined ví $x = \frac k 2 \pi; k$ even or in other words $x = \frac n \pi$.

answered Aug 8, năm nhâm thìn at 22:18

$\endgroup$

$\begingroup$

When you have this: $$-\tan^2{x} \cdot \frac{2\tan{x}}{1-\tan^2{x}} = \frac{2\tan{x}}{1-\tan^2{x}}$$

It it incorrect vĩ đại have this next: $$-\tan^2{x} \cdot \frac{2\tan{x}}{1-\tan^2{x}} \cdot \frac{1-\tan^2{x}}{2\tan{x}} = 1$$

Just lượt thích when you have (for example) $3 \cdot 2x = x$, it is incorrect vĩ đại then divide both sides by $x$ and kết thúc up with $6 = 1$ (you lose the solution $x = 0$ when you vì thế this).

The correct next step is vĩ đại subtract $\dfrac{2\tan x}{1-\tan^2 x}$ from both sides and continue as follows:

\begin{align} -\tan^2{x} \cdot \frac{2\tan{x}}{1-\tan^2{x}} &= \frac{2\tan{x}}{1-\tan^2{x}}\\[0.3cm] -\tan^2{x} \cdot \frac{2\tan{x}}{1-\tan^2{x}} - \frac{2\tan{x}}{1-\tan^2{x}} &= 0\\[0.3cm] \frac{2\tan{x}}{1-\tan^2{x}} \left(-\tan^2 x - 1\right) &= 0 \end{align} So then we have the following two equations: $$\frac{2\tan x}{1-\tan^2 x} = 0 \qquad \text{or} \qquad -\tan^2 x - 1 = 0$$

The first equation is trivial. The second one is perhaps even more ví because the second one has no solutions.

answered Aug 8, năm nhâm thìn at 22:20

$\endgroup$

$\begingroup$

Why vì thế you have an error?

If you divide though (by $\frac {2\tan x}{1-\tan^2x} = \tan 2x$ in this case), you must kiểm tra that whatever you are dividing by does not equal 0. Or, make a note vĩ đại review the case that it does.

$\tan^2 x = -1$ or $\frac {2\tan x}{1-\tan^2x} = 0$

An alternatives you could have taken:

$0 = (1 - \tan^2 x)\frac {2\tan x}{1-\tan^2x}\\ 0 = \sec^2 x \tan 2x$

But you are working a little bit too hard.

$\tan x = \frac {\tan x + \tan 2x}{1-\tan x\tan 2x}$

Rather kêu ca expand $\tan 2x$ right now, lets simplify a bit first

$\tan x- \tan^2 x \tan 2x = \tan x + \tan 2x\\ 0 = \tan 2x (1+\tan^2 x)\\ 0 = \tan 2x(\sec^2 x)$

Now, let's be even a little bit stupider.

$\tan x = \tan (x + k\pi)\\ \tan (x + k\pi) = \tan 3x\\ \tan^{-1}(\tan (x + k\pi)) = \tan^{-1}(\tan 3x)\\ x + k\pi = 3x\\ 2x = k\pi$

answered Aug 8, năm nhâm thìn at 22:23

$\endgroup$