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Solve the equation

$$\cot x-\tan x=4\cot(2x)$$

for $0^\circ<x<360^\circ$

My attempt,

$$\frac{1}{\tan x}-\tan x=4(\frac{1}{\tan 2x})$$

$$\frac{4(1-\tan^2x)}{2\tan x}-\frac{1}{\tan x}+\tan x=0$$

$$2-2\tan^2 x-1+\tan^2x=0$$

$$1-\tan^2x=0$$

$$\tan x=\pm1$$

$$x=45^\circ,135^\circ,225^\circ,315^\circ$$

So I've checked out with Desmos which I got

enter image description here

It shows that the answer is $45$ and$135$.

My questions:

1)Why my $225^\circ$ and $315^\circ$ are not included in the graph? Are they incorrect?

2)If I substitute the answers back to the equation, for example,

$$4\cot (2 \cdot 45)$$ which is undefined. Why?

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    $\begingroup$ Cot (90)=0 no?... $\endgroup$ – Archis Welankar Jun 1 '17 at 15:52
  • $\begingroup$ I guess I took cot x=1/tanx. It's defined when I take cotx=cosx/sinx. Why? $\endgroup$ – Mathxx Jun 1 '17 at 15:55
  • $\begingroup$ Tan (90)=\infty $ so either way its zero . Is that your doubt $\endgroup$ – Archis Welankar Jun 1 '17 at 15:58
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Following a thorough remark by @lab bhattacharjee, I realized that my initial answer has been influenced by the way you have transformed the equation.

Here is a very simple way to handle the issue.

Let us write the initial equation in the following way:

$$\underbrace{\cot x-\tan x}_{2\cot 2x}-4\cot(2x)=0,$$

which is plainly equivalent to... $\cot(2x)=0,$

with solutions $2x=(2k+1)90^{\circ}, \ \ k=0,\cdots 3$, that is to say:

$$x=45°, \ \ \ \ 3 \times 45°, \ \ \ \ 5 \times 45°, \ \ \ \ 7 \times 45°$$

Remark: what are the equations of the blue and red curves you have plotted, in order to understand where the domains on which we are looking for solutions have been modified ?

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  • $\begingroup$ If $\cot45^\circ-\tan45^\circ=2\cot(2\cdot45^\circ)$ what is wrong with $$\cot(180+45)^\circ-\tan(180+45)^\circ=2\cot2((180+45)?$$ $\endgroup$ – lab bhattacharjee Jun 4 '17 at 16:01
  • $\begingroup$ @lab bhattacharjee You are fully right. I realize now that I have answered to the OP concerning a transformed equation where poles have been added. This has given me the opportunity to reconsider the problem and rewrite a very simple solution. $\endgroup$ – Jean Marie Jun 4 '17 at 20:58
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$$\cot x-\tan x=2\cdot\dfrac{\cos^2x-\sin^2x}{2\sin x\cos x}=2\cot2x$$

So, we need $2\cot2x=4\cot2x$ assuming $\sin x\cos x\ne0$

$\iff\cot2x=0\iff\cos2x=0$

$\implies2x=(2n+1)90^\circ$ where $n$ is any integer

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Looking at the Desmos graph (and working the equation out for myself), I discovered $45°$ and $225°$ (not $135°$) are the two solutions within $x \in (0°, 360°)$. Due to the quadratic nature of $\tan^2 x = 1$, the solutions of $135°$ and $315°$ are extraneous. (The general solution for all $n$ are $45° + 360° \cdot n$ and $225° + 360° \cdot n$).

Checking with $135°$ on each side: $$\cot 135° - \tan 135° = -1 - (-1) = 0$$ but $$4 \cot (4 \cdot 135°) \rightarrow 4 \cot 180°$$

However, $\cot 180°$ is undefined, so there is no solution. (It will yield the same answer for $315°$.)

In the graph, there is no graph through $135° \text {and} \ 315°$, - only through $45° \text {and} \ 225°$.

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