1
$\begingroup$

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?

$\endgroup$
3
  • 2
    $\begingroup$ 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
1
$\begingroup$

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 ?

$\endgroup$
2
  • $\begingroup$ 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
2
$\begingroup$

$$\cot x-\tan x=2\cdot\dfrac{\cos^2x-\sin^2x}{2\sin x\cos x}=2\cot2x$$

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

$\endgroup$
0
$\begingroup$

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 $45° \text {and} \ 225°$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.