When I try to solve the above equation using $t=\tan(x/2)$, it reduces to a linear equation in t giving $x=\pi/2 + 2n\pi$. However, I completely miss the solution $x=(2n+1)\pi$ this way. My test book says that all "quadratic" equations have an infinite solution when the quadratic term cancels so $\tan(x/2)=\pm\infty$ is also a valid solution. What's actually going on here?

I am ready aware that the above equation can be solved by using the double angle formula or expressing it as a single sine or cosine so pointing our the error rather than providing another solution would be appreciated

  • 1
    $\begingroup$ The question changed from $\cos x-\sin x=1$ to $\sin x-\cos x=1$. Which one is it? $\endgroup$ Commented Jun 1, 2020 at 13:18
  • $\begingroup$ I am very sorry for the mess up. Its sinx-cosx=1 $\endgroup$ Commented Jun 1, 2020 at 13:20
  • $\begingroup$ Refrain from changing questions once answers are received. $\endgroup$ Commented Jun 1, 2020 at 13:21
  • $\begingroup$ math.stackexchange.com/questions/618192/… $\endgroup$ Commented Jun 1, 2020 at 18:04

4 Answers 4


I thought it was worth making this answer to explain what your text was (maybe) referring to, with regard to "solutions at infinity" and how it ties in here.

The general quadratic equation is $ax^2 + bx + c = 0$. Usually, $a \ne 0$ is an assumed condition. However, let's see what happens if we disregard that condition.

The general solution is usually represented as $\displaystyle x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$. Consider what happens as we let $a \to 0$ while $b,c \ne 0$. One root becomes the inderminate form $\frac 00$ (which we'll address soon), while the other root becomes $\frac ba \to \pm\infty$ (depending on the signs of $a$ and $b$). This is (probably) what your text is talking about when they refer to a "root at infinity".

What about the other root of the form $\frac 00$? Well, we can't make much headway using the standard formula. However, we can tranform the formula using one of Vieta's formulas for the product of roots (which will be equal to $\frac ca$). Using that, we get the alternative quadratic formula $\displaystyle x = \frac{2c}{-b \pm \sqrt{b^2-4ac}}$. When you apply the same conditions to this formula, you get one root as $\frac{2c}{0} = \pm \infty$ (depending on the sign of $c$), and the other root as $-\frac cb$, which is very much a defined real number.

What's going on here? Obviously, when you think about it, if you drop (or "cancel" for whatever reason) the lead term, you make the quadratic a simple linear equation $bx + c = 0$. This has only one solution, namely $x = -\frac cb$. We have managed to recover this (real, defined) root as the second value found with the "alternate" quadratic formula. It is consistent with the indeterminate form $\frac 00$ found by the first method because that indeterminate form may actually assume any value, including $0, \pm \infty$ and any finite value (in this case, the latter possibility holds).

So we have a defined root $x = -\frac cb$ and a "root at $\pm \infty$".

Does any of this make any sense? Not in usual algebra and analysis, because what I wrote is highly non-rigorous. A quadratic is a polynomial of degree $2$ and we're usually focussed on finding real (or sometimes complex) solutions to these. Infinity is not a real number (neither is it a complex number). It does not lie on the real line. However, it is my understanding that it is possible to rigorously extend the real numbers to so-called "hyperreal numbers" and "surreal numbers" that include the concept of infinity on the number line, but I don't know enough about this, so I won't talk about them further. The bottom line is that the generally accepted method of solution of quadratics only admits of defined real and complex solutions, and solutions like "roots at infinity" are treated as absurdities.

But we have to see how all this ties in to your problem. When one deals with infinities, generally limits have to be considered. The transform you used (Weierstrass substitution) was $\displaystyle \sin x = \frac{2t}{1+t^2}$ and $\displaystyle \cos x = \frac{1-t^2}{1+t^2}$. Let's see what happens when we let $x = (2n+1)\pi,\ n\in\mathbb{Z}$. In this case, $t$ will be unbounded, and the right hand sides will both become indeterminate forms $\pm \frac{\infty}{\infty}$. However, if we used the theory of limits to find the limiting behaviour of the functions, we get:

$\displaystyle \sin (2n+1)\pi = \lim_{t \to \pm \infty} \frac{2t}{1+t^2} =\lim_{t \to \pm \infty} \frac{2}{\frac 1t +t} = \lim_{t \to \pm \infty} \frac 2{t} = 0$


$\displaystyle \cos (2n+1)\pi = \lim_{t \to \pm \infty} \frac{1-t^2}{1+t^2} =\lim_{t \to \pm \infty} \frac{\frac 1{t^2} - 1}{\frac 1{t^2} +1} = -1$

so we have successfully "recovered" the usual relations.

But when you use the $t = \tan\frac x 2$ substitution without considering the limiting behaviour when $t$ is not defined for certain parts of the domains where the original functions ($\sin x, \cos x$) are well-defined, that's when you run into problems. In your case, as you work through the algebra, you'll find a step where you get a quadratic on the LHS and another on the RHS, and the two lead terms "cancel" to give a linear equation. By reducing that to a linear equation, you lose the roots that occur when $t$ is undefined. However, if you take the alternative view of the equation as being the limiting case as $a \to 0$ like I explained above, you'll be able to recover the solutions $t = \pm \infty$, which will give the "missing" solutions.

The bottom-line is that I consider this a poor technique to solve the problem because of the difficulties it presents. As others have mentioned, it is a basic rule of thumb that when you make a substitution it behoves you to ensure that your substituted function(s) is/are properly defined for the same domain as the parent function(s). This is clearly a case where this was not done properly or rigorously, so the method seems to fail without going through a lot of convolutions (and non-rigorous reasoning).

I'm not sure at what level your text is targetted, nor the exact phrasing used, but I consider the mention of "roots at infinity" etc. to be very poor form. Introducing concepts like these to students who are not yet familiar with calculus or limits is bound to cause confusion and lead in turn to them forming bad habits of their own, in my view. On that basis, I consider at least this part of your text to be very poorly written.

On the whole, it's better to use another method of solving the equation. The conversion to either of the $R\sin(x\pm \alpha)$ or $R\cos (x\pm \alpha)$ forms is my favourite as it is a one-to-one transform and it gives all the roots every time (and no extraneous roots), but you can also square both sides of the equation. Keep in mind that squaring (or raising any equation to an even power) will introduce "extraneous roots", which you must eliminate by checking against the original equation. This ties in with the larger theme of the method you use determining exactly how simple and direct the solution is.

  • $\begingroup$ Thankyou so much for your answer. This pretty much perfectly answers all the questions I had. The “solutions at infinity” were mentioned in “The core course for A level” by L. Bostock and believe it or not but is regarded as one of the best texts for A level mathematics. But you are right; even though i do have a non rigorous knowledge of limits, they were not at all mentioned. $\endgroup$ Commented Jun 1, 2020 at 15:07
  • $\begingroup$ The problem i have with expressing as a single sine or cosine is that the solutions, at least for some equations, aren’t as neat when expressed in exact form. But i guess its better than actually losing solutions due to carelessness $\endgroup$ Commented Jun 1, 2020 at 15:09
  • $\begingroup$ @OVERWOOTCH Glad to help. I have actually read books co-written by Bostock (and Chandler), namely "Pure Mathematics" (two volumes) for my 'A' levels, but I don't recall them containing anything of such a ghastly non-rigorous nature. $\endgroup$
    – Deepak
    Commented Jun 1, 2020 at 15:09
  • $\begingroup$ But that was around $28$ years ago, and I was probably less "mathematically mature" then, so who knows? :) $\endgroup$
    – Deepak
    Commented Jun 1, 2020 at 15:15

Other than using the substitution, I think there is another way to solve it. If you square both sides, we get $$(\cos x-\sin x)^2=1,$$ which gives $$\cos^2 x+\sin^2 x-2\cos x\sin x=1.$$ Since $\cos^2x+\sin^2x=1$, we have $\cos x \sin x=0$, which implies $\cos x=0$ or $\sin x=0$. From these, we get $x=k\pi+\pi/2$ and $x=k\pi$ where $k$ are integers.

  • $\begingroup$ Good answer. Even simpler is $\sin(2x)=0$ which means any half-integer multiple of $\pi$ is a solution. So $\pi/2, \pi, 3\pi/2,$ etc. $\endgroup$
    – K.defaoite
    Commented Jun 1, 2020 at 13:11
  • $\begingroup$ Im not looking for another method, as ive mentioned. Where is the substitution failing? $\endgroup$ Commented Jun 1, 2020 at 13:14
  • $\begingroup$ As pointed out to me in an answer I've deleted, even $k$ does not satisfy your first answer ($x=k\pi+\pi/2$). $\endgroup$ Commented Jun 1, 2020 at 13:17
  • $\begingroup$ I messed up the equation in a hurry. Im very sorry. Its sinx-cosx=1 $\endgroup$ Commented Jun 1, 2020 at 13:22

You forgot that $\tan(x/2)$ is not defined for $x/2=(2n+1)\pi/2$ i.e. $x=(2n+1)\pi$. Check those values and they turn out to be solutions too.

  • $\begingroup$ I am aware of that. My question is why should i even consider the undefined values in the first place. $\endgroup$ Commented Jun 1, 2020 at 13:24
  • $\begingroup$ Both $\sin x$ and $\cos x$ are defined for $x=(2n+1)\pi$. Your substitution restricts your domain. $\endgroup$ Commented Jun 2, 2020 at 22:00

If $x$ is such that $t=\tan(x/2)$ is undefined, then the original equation is perfectly well-defined, but the substitution is illegal. So it's a good idea to check if these satisfy the original equation.

  • $\begingroup$ Does that mean I should refrain from using the t substitution in solving such equations (since i am solving for x and won't know if tan(x/2) is undefined before hand) $\endgroup$ Commented Jun 1, 2020 at 13:26
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    $\begingroup$ @OVERWOOTCH Worry not, you can still use it! But just also check situations in which it's undefined. Your mental logic should be: "Either $\tan(x/2)$ is undefined, which gives a small number of possible solutions for $x$ which I can check by hand, or $\tan(x/2)$ is defined, in which case I make the (now valid) substitution and find any other solutions." $\endgroup$
    – BallBoy
    Commented Jun 1, 2020 at 13:29

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