Just simplify everything, right?

Geometry Level 4

In the domain $[ 0^ \circ, 360 ^ \circ ],$ how many solutions are there to $\frac {\tan 2x- \tan x}{1+\tan 2x\tan x}=1?$

0 1 2 3

2 solutions

Suvajyoti Barman
Mar 20, 2017

Most recognize that the LHS is the difference formula for $\tan ( 2x -x )$ , and so think that we have solutions $\tan x = 1$ which gives $x = 45 ^ \circ, 225 ^ \circ, 405 ^ \circ , \ldots$ .

However, notice that when $\tan x = 1$ , we have $\tan 2x$ is undefined. Thus, the original expression would be undefined (in both the numerator and the denominator).

Hence, the equation has 0 solutions.

Moderator note:

Be aware of gaining/losing solutions when we square/root equations, multiply/divide by 0, etc. Often when we write down our series of steps, we are careless and forget to jot down the necessary conditions for us to get a double-sided implication.

this is the most common mistake i always do. forget to check extancious roots

Nivedit Jain - 4 years, 2 months ago

You are not alone. Only 30% got it correct.

Calvin Lin Staff - 4 years, 2 months ago

ya that is the reason it is in advance section. Right??

Nivedit Jain - 4 years, 2 months ago

Now it is only 23%....!!

Aaghaz Mahajan - 3 years, 1 month ago

But when u take these values undefined value will come bcz tan 2x will be tan 90 and hence no solution.

Suvajyoti Barman - 4 years, 2 months ago

But when you take the limit of the function it tends to 1. Therefore there can be 2 solutions 45 and 225

Sai Venkatesh - 4 years, 2 months ago

The limit of the function may equal to 1, but that doesn't not imply that the value at that point is equal to 1.

For example, even though $\lim_{x \rightarrow 0 } \frac{ x}{x} = 1$ , we do not say that $\frac{ 0 } { 0 } = 1$ .

Calvin Lin Staff - 4 years, 2 months ago

we can take (tan 2x) common then we get [(tan 2x)(1-(tanx/tan2x))]/[(tan2x)((1/tan2x)+tanx)] which on applying limit x tending to (pi)/4 or 5(pi)/4 we get [(1-0)/(1+0)] =1

Preetam Kandula - 4 years, 2 months ago

@Preetam Kandula – What are the condtions under which we can "take tan 2x common"? Note that you are "dividing by tan 2x", and so we have to watch out for what this value is.

For example, if we wanted to solve $\frac{ 1}{x-1} \times x = \frac{1}{x-1}$ , are we allowed to "take $\frac{1}{x-1}$ common" and cancel it and claim that $x = 1$ ?

Calvin Lin Staff - 4 years, 2 months ago

I took the view that the function above was continuous - except at 135 and 315. It seems rather stingy to disallow the two values at 45 and 225 just because tan 2x is infinite

Malcolm Rich - 4 years, 2 months ago

@Malcolm Rich – It is not "stingy to disallow just because X is infiinte".

Note that $\tan 90 ^ \circ$ is "undefined", as opposed to "infinite". Saying that $\frac{ \tan 90^ \circ - 1 } { \tan 90 ^ \circ + 1 } = 1$ is analogous to saying that $\frac { \frac{ 1}{ 0 } } { \frac {1}{0} } = 1$ .

Calvin Lin Staff - 4 years, 2 months ago

@Calvin Lin – Let me ponder that awhile. The example you've given is essentially (sinx-cosx)/(sinx+cosx) which is a well-behaved function.

Malcolm Rich - 4 years, 2 months ago

@Malcolm Rich – To clarify, the example (first equation) that I gave is substituting in $x = 45 ^ \circ$ and then $\tan 45 ^ \circ = 1$ in the problem statement. IE It is what you are claiming.

What I am claiming, is that your statement is equivalent to $\frac{ 1/0 } { 1/0} = 1$ . They are not exactly identical, but both numerators and denominators are undefined, which is my main point.)

Note that $\frac{ \sin x - \cos x } { \sin x + \cos x }$ is not a "well behaved function", specifically when $\sin x + \cos x = 0$ . Analagously, that's like saying "The function $\frac{ 1 } { x }$ is well behaved, even at $x = 0$ .

Furthermore, not that we cannot multiply throughout by $\frac{ 0 } { 0 }$ , which occurs when $\cos x = 0$ , for the "essentially" aspect.

Calvin Lin Staff - 4 years, 2 months ago

@Calvin Lin – Would you have the same problem with x^2/x?

Malcolm Rich - 4 years, 2 months ago

@Malcolm Rich – Depends on how you write it, and what you are referring to.

The issue with $\frac{ x^2 } { x }$ at $x =0$ is $0 / 0$ , which is undefined.

As an aside, note that we cannot multiply by $\frac{ f(x) } { f(x) }$ at the points where $f(x) = 0$ . IE it is not fair to say that $\frac{ 1 } {1}$ is the same as $\frac{ x} { x }$ .

Calvin Lin Staff - 4 years, 2 months ago

@Calvin Lin – To say that x^2/x does not "exist" at x=0 seems to me illogical. The function f(x)=(x^2)/x is clearly equivalent to g(x)=x. Your reasoning would imply that the continuous function f(x)=g(x) exists for all x except x=0.

Malcolm Rich - 4 years, 2 months ago

@Malcolm Rich – At this point in time, we agree that we disagree about " $\frac{ x^2 } { x}$ is not defined at $x = 0$ ". There isn't too much that I can say that would convince you that 0/0 is undefined.

Note: when graphing rational equations , we say that the function $\frac{ x^2 } { x }$ has a hole at $x = 0$ . Yes, we would love to fill it in "by continuity", but we're not allowed to. Of course, if we had other "regularity / real world" conditions, then go ahead and fill it in.

Note: Solving $f(x) = g(x)$ is not equivalent to solving $\frac{ f(x) } { g(x) } = 1$ . Speficailly, we have to consider the edge case of when $g(x) = 0$ , because we are dividing by 0.

Calvin Lin Staff - 4 years, 2 months ago

very tricky... =(

Reygan Dionisio - 4 years, 2 months ago

Haha yeah just realised it

Noel Lo - 4 years, 2 months ago

Ha Ha 😂😂... I also made the same mistake... Didnt took tan2x into account...

Sampad Kar - 4 years, 2 months ago

I don't get why we ask about tan2x. If the LHS reduces to tanx from tan(2x-x) doesn't that mean the two solutions work?

Dominic Dharam - 4 years, 2 months ago

If y = 1 is the same as y = (x-1)/(x-1), then is y = 1 undefined at x = 1?

Jim Laverty - 3 years, 3 months ago

The first 2 equations are not the same. For $y = \frac{x-1}{x-1}$ , it is undefined at $x = 1$ because of the division by 0. If you graphed it, there will be a hole.

Note that in this question, the error is because one of the terms is undefined (as opposed to having a division by 0).

Calvin Lin Staff - 3 years, 3 months ago

There was a similar question asked in AIEEE

Gitali Dutta - 1 year, 4 months ago

let apply Tan{a-b} formulae then it come tanx so it can have 45 and 225 as its solutins

Lakshya Chhaangani - 4 years, 2 months ago

Does putting 225 degrees really work?

Agnishom Chattopadhyay - 4 years, 2 months ago

Naren Bhandari
Mar 30, 2017

Solving the equation on LHS gives $\tan x = 1$ In General Form we can write it as $\tan x = \tan (n\pi+ \frac{\pi}{4})$ so Possible solutions for $x = 45^\circ, 225^\circ , 405^\circ .... for \space n =0,1,2....$ which shows that there is no solution for x in the interval $[0^\circ, 360^\circ]$

Could you explain how you went from "possible solutions are $x = 45^\circ, 225^\circ \ldots$ to "there is no solution for x..."? We can see that $45^\circ$ and $225^\circ$ belong to $[0^\circ, 360^\circ]$ .

Pranshu Gaba - 4 years, 2 months ago

Since a solution of the form where x = 45, 225,.... would give corresponding 2x = 90, (360+90)... and so on which is not defined for tan2x. Thus these are not acceptable solutions.

Unnati Akhouri - 4 years, 2 months ago

But applying limit on these values (x=45 degrees,225 degrees), we do get the result as 1. Initial equation is also of the form infinte/infinte. So, why is applying limit wrong?

Aeshit Singh - 4 years, 2 months ago

@Aeshit Singh – its not wrong...as stated above by Brilliant staff ..even though the limit of x tends to 0 (x/x)is equals to 1...we don't say that 0/0=1.

Suvajyoti Barman - 4 years, 2 months ago

Just simplify everything, right?

2 solutions

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