My Overrated Problem

Geometry Level 3

How many real numbers $x$ in radian satisfy the equation

$\cos x\tan x+\sin x\cot x=1?$

0 1 2 3 4 Infinitely many solutions

3 solutions

Paul Ryan Longhas
Jan 26, 2015

cosxtanx + sinxcotx = 1 => sinx + cosx = 1 => 2sinxcosx = 0. So, sinx = 0 and cosx = 1 or cosx = 0 and sinx = 1. But, If sinx = 0 --> cotx is not defined. Also, If cosx = 0 then tanx is not defined. Therefore, x has no Real solution.

I do not understand why the solutions x=2pi n, being n an integer are dismissed, besides the equation sinx+cosx=1 has also the solution x=1/2(4pi n+pi/2) you can check these results at :

https://www.wolframalpha.com/input/?i=sin+x+%2B+cos+x+%3D+1&lk=3

Where you can appreciate the equation has infinite solutions

Mariano PerezdelaCruz - 6 years, 4 months ago

I tend to agree with you, Mariano. I think that it depends on how we define $\tan(x)$ and $\cot(x)$ , and whether we can look at $\cos(x)\tan(x)$ and $\sin(x)\cot(x)$ as "whole" entities rather than breaking them down into component parts.

With the definition $\tan(x) \equiv \frac{\sin(x)}{\cos(x)}$ we can infer that $\sin(x) \equiv \cos(x)\tan(x)$ . Likewise, with $\cot(x) \equiv \frac{\cos(x)}{\sin(x)}$ we can infer that $\cos(x) \equiv \sin(x)\cot(x).$ The equation then does become $\sin(x) + \cos(x) = 1$ , which clearly has an infinite solution set. My only concern here is whether there is difference in the way we can deal with equivalences, (definitions), and statements of equality, a concern that I will have to investigate further.

If we break the two products into component parts, however, we have two cases of the indeterminate form $0*\infty$ at the values you mention, which would mean that there are no solutions.

So I think that there has been a subtlety lost here in saying that either " $0$ " or "infinitely many" is the correct answer, as there is an issue of interpretation that needs to be dealt with. For what it's worth, WolframAlpha does seem to agree with our shared answer, namely "infinitely many", but WA is not a definitive source.

Brian Charlesworth - 6 years, 4 months ago

I disagree, though I can see where you're coming from.

With the definition $\tan x = \frac{ \sin x } { \cos x}$ , you should also bear in mind that $\tan x$ is undefined for $x = (2n+1) \pi$ . Furthermore, when multiplying throughout, we have to take care when we multiply by $\cos x = 0$ , and ensure that we don't do $0 \times \infty$ , which is undefined like you pointed out,

For these reasons, the equation does not (strictly) hold true at $x = (2n + 1) \pi$ .

Calvin Lin Staff - 6 years, 4 months ago

@Calvin Lin – Fair enough, but the "equivalence/equality" dichotomy still intrigues me. I deliberately used the $\equiv$ symbol in $\tan(x) \equiv \frac{\sin(x)}{\cos(x)}$ to distinguish it from an equality. With the = sign as you have it then I can see $\sin(x) = \cos(x)\tan(x)$ only when $\cos(x) \ne 0$ , but with the equivalence sign I just wonder if $\tan(x) \equiv \frac{\sin(x)}{\cos(x)} \Longrightarrow \sin(x) \equiv \cos(x)\tan(x)$ for all $x$ , in which case we can replace $\cos(x)\tan(x)$ with $\sin(x)$ in the original equation without reservations.

I'm not going to dispute the posted answer of $0$ , (I think that almost all of the questions I get wrong on Brilliant are of the multiple-choice variety because there always seems to be some trick or nuance that is being employed, and you only get one kick at the can), but at least it has raised an interesting subtlety that I now need to resolve for myself.

P.S.. I do tend to overthink things, huh? :)

Brian Charlesworth - 6 years, 4 months ago

cosxtanx + sinxcotx = 1 => sinx + cosx = 1. So x=2pi n or x=1/2(4pin+pi/2). But, if x = 2pi n -> cotx is not defined and also if x=1/2(4pin+pi/2) -> tanx is not defined. So, x= 2pi n or x=1/2(4pin+pi/2) is the solution only for sinx + cosx=1 but not in cosxtanx + sinxcotx=1.

Paul Ryan Longhas - 6 years, 4 months ago

I do not agree with you may be you can say that cotx is not define for x=0, but remenber we are considering (sinx)*(cosx)/(sinx) which limit is 1 for x tending to 0 same argument for pi/2.

Mariano PerezdelaCruz - 6 years, 4 months ago

@Mariano PerezdelaCruz – While the limit is defined, the value does not exist for $x = 0$ .

For example, we do not say that 0 is a solution to the equation $\frac{x}{x} = 1$ .

Calvin Lin Staff - 6 years, 4 months ago

@Calvin Lin – I dissent with your opinion just verify in wolfram alpha the very same equation

https://www.wolframalpha.com/input/?i=cosxtanx+%2B+sinxcotx+%3D+1

and they give you the roots

our function is cos(x)tan(x) + sin(x)/tan(x). For now we focus in the last term which renders one of the solutions. The value of our function at x=0 is the limit of the function at that point isn’t? the first term is obviously 0 the second one is 1 since the quotient sin(x)/tan(x) ; x/tan(x) or sin(x)/x is 1 arc, sin and tan became equals in the proximity of x=0.

Mariano PerezdelaCruz - 6 years, 4 months ago

@Mariano PerezdelaCruz – Do not be too reliant on external calculation aids, if you are unaware of their limitations. While Wolfram Alpha is indeed a powerful tool, there are several things which I would not use it for. In particular, in this case it made simplifying assumptions when solving the problem, and does not verify that these assumptions hold.

When we're solving an equation, we do not necessarily need to consider what happens in a neighborhood of a point. For example, if $f(x)$ is defined to be 1 for integers and 0 otherwise, would you say " $x = 0$ is a solution to $f(x) = 0$ because in a neighborhood of 0 (excluding 0) we have $f(x) = 0$ ."

By your argument, is it true that $\frac{ 0} { 0} = 1$ by considering $\frac{x}{x}$ in proximity of $0$ ? The value in a neighborhood has no implication on the value of a point if the function is not continuous at the point.

Calvin Lin Staff - 6 years, 4 months ago

@Calvin Lin – I still wonder, though, if there are some nuances to be dealt with here from the standpoint of definitions. I've outlined my concerns below. The $\frac{0}{0}$ discussion seems to be coming up with great frequency of late. :)

Brian Charlesworth - 6 years, 4 months ago

Yes it's a very common mistake we make we don't see that the roots which we are getting are make the given function defined or not.

What mistake you are doing is that you are considering the given function as sinx + cosx., but is incorrect.

Try this another same kind

U Z - 6 years, 4 months ago

Wow ! I got it wrong because I didn't see the phrase " REAL solution" :v

Trung Đặng Đoàn Đức - 6 years, 2 months ago

Thomas James Bautista
Feb 12, 2015

If we have $\tan x$ and $\cot x$ in the equation, then $\sin x$ and $\cos x$ cannot be 0.

If we let $\tan x = \frac{\sin x}{\cos x}$ and $\cot x = \frac{\cos x}{\sin x}$ , we can simplify the equation and have $\sin x + \cos x=1$ .

Squaring the whole equation, we get $\sin^2x+2\sin x\cos x+\cos^2x=1$ .

Substituting the Pythagorean identity $\sin^2x+\cos^2x=1$ , we get $1+2\sin x\cos x=1$ , or $2\sin x\cos x=0$ .

According to the zero-product property, $\sin x=0$ or $\cos x=0$ , which contradicts our first statement.

Therefore, there are no solutions to the equation.

Figel Ilham
Feb 9, 2015

Since $\cos(x)\tan(x)=\sin(x)$ for all real with $x\neq \frac{\pi}{2} +\pi k, k\in \mathbb{Z}$ and $\sin(x)\cot(x)=\cos(x)$ for all real with $x\neq \pi k, k\in \mathbb{Z}$ , we conclude that

$\cos(x)\tan(x)+\sin(x)\cot(x)=\sin(x) + \cos(x), x\neq \frac{\pi k}{2}, x\in \mathbb{Z}$

Squaring both sides,

$(\sin(x)+\cos(x))^2=1^2$

$\sin^2(x)+\cos^2(x)+2\sin(x)\cos(x)=1$

Since $\sin^2(x)+\cos^2(x)=1$ , we write:

$1+2\sin(x)\cos(x)=1$

$\sin(2x)=0$

$2x=\arcsin(0)=\pi k, k\in \mathbb{Z}$

$x=\frac{\pi k}{2}, k \in \mathbb{Z}$

contradicts what we have defined the value of $x$ before. So, there are no solutions!

That's what I thimk, too!

Margarida Pereira - 5 years, 7 months ago

Best solution.

Márcio Gomes - 5 years, 5 months ago

My Overrated Problem

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