\begin{align*} \sec x\cdot\cos5x+1&=0, \qquad 0<x<2\pi\\ \frac1{\cos x}\cdot\cos5x+1&=0\\ \cos5x+\cos x&=0\\ 2\cos3x\cos2x&=0 \qquad [\because \cos a+\cos b=2\cos\left(\frac{a+b}2\right)\cos\left(\frac{a-b}2\right)]\\ \end{align*}

$$ \begin{array}{cc} \Rightarrow \cos3x=0 & \Rightarrow \cos2x=0\\ \cos3x&=\cos\frac\pi2 & \\ \Rightarrow 3x=(2n+1)\frac\pi2, n\in\mathbb Z & \\ x=(2n+1)\frac\pi6, n\in\mathbb Z & \end{array} $$ Now for $n=1$ \begin{align*} x&=\frac\pi2\\ \sec x\cdot\cos5x+1&=0, \qquad (\text{for }x=\frac\pi2)\\ \sec\frac\pi2\cdot\cos\frac{5\pi}2+1&=0\\ \frac{\cos(2\pi+\frac\pi2)}{\cos\frac\pi2}&=0\\ \Rightarrow \frac{\cos\frac\pi2}{\cos\frac\pi2}+1&=0 \qquad \Rightarrow 2=0 \text{(Possible?)}\\ \Rightarrow \frac{0}{0}+1&=0 \end{align*} Or can we say that $\frac00$ (which is undefined) is $-1$?

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    You cannot evaluate $\sec(x)$ at $x=\pi/2$. It is not defined. –  Sep 18 '16 at 13:41
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    No, it's not the way it works. You may never divide the members of an equation by zero. The original equation implies that $\cos x\ne0$. –  Sep 18 '16 at 13:42
  • @YvesDaouts then why I am getting $cosx=0$ ? Where did I did mistake? – Fawad Sep 18 '16 at 13:59
  • I have retyped the text from your picture. (Whenever possible, text is preferable to images for obvious reasons.) Try to check it and edit it further to the form you are satisfied with. – Martin Sleziak Oct 08 '16 at 10:02
  • We can't say it equals to -1 or any other number. That's the point of calling an expression undefined, isn't it? – 9Algorithm Oct 08 '16 at 10:04
  • @9Algorithm but for this case putting it as -1 hold the condition! So why can't we? – Fawad Oct 08 '16 at 10:06
  • @Ramanujan No, it doesn't. You can't divide by zero, this operation is undefined. So your expression can't be evaluated. There's no flaws in mathematics, they would be if one could divide by zero in $R$. – 9Algorithm Oct 08 '16 at 10:13

3 Answers3


No, you can't say $0/0=-1$, because $0/0$ cannot be equal to anything, being undefined.

More easily: your equation has $\sec x$, so you have to exclude $\pi/2+k\pi$ ($k$ integer) from the solutions, because the expression you have is undefined for those values.

Then, using the sum-to-product formulas you correctly get $$ 2\cos3x\cos2x=0 $$ This leads to two families of solutions.

First family: $\cos3x=0$

This means $3x=\frac{\pi}{2}+k\pi$, so $$ x=\frac{\pi}{6}+k\frac{\pi}{3} $$ with $k$ integer not divisible by $3$.

Second family: $\cos2x=0$

This means $2x=\frac{\pi}{2}+k\pi$, so $$ x=\frac{\pi}{4}+k\frac{\pi}{2} $$ with $k$ integer not divisible by $2$.

Final note

The values we exclude from the solutions above are not solutions of the equation; plugging them in produces undefined expressions and you can't draw conclusions about $0/0$ (which is not defined) from something which is undefined as well.

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To evaluate $\frac{0}{0}$, you must ask yourself the question "How many zeroes must be added together to arrive at zero?"

The answer to this is, of course, that any number of zeroes added together, will still give you zero. And that means the answer is pretty much anything you want.

For that reason, in general algebra it is regarded as an operation which is not permitted. There is some pretty deep maths surrounding this question but if we draw upon some very complicated calculus, there is a stronger argument that $\frac{0}{0}=0$ than any other number, but I will have to dig out the argument... and it is not simple!

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  • Then what can we say when we get answer as $\frac{0}0$ from our solving of answer? – Fawad Oct 08 '16 at 10:30
  • @Ramanujan: That we have made an error somewhere. – celtschk Oct 08 '16 at 10:59
  • Can you spot any error there? – Fawad Oct 08 '16 at 11:00
  • @Ramanujan: You multiplied both sides of the equation with $\cos x$, without adding the additional condition that $\cos x\ne 0$. For $\cos x=0$ that is not an equivalence transformation, and may (and in your case, does) add more solutions. In particular, the solution you use for your argument is *not* a solution of the original equation. – celtschk Oct 08 '16 at 11:44
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    @Ramanujan What this means is that the information about $x$ contained within the original equation, has been lost in your algebra. So the answer is "pretty much anything you want". If I say $x=5$ and I allow myself to multiply both sides by $0$ then I have $0x=5\times0=0$ and if I then allow myself to divide both sides by zero then I have $0x/0=0/0$, so $x$ can be anything I want. The restrictions on $x$ contained within the original equation have been lost. When making a jump from one equation to the next, it is helpful to ask is it true if you insert a $\iff$ in-between sequential lines. – samerivertwice Oct 10 '16 at 09:50

When we solve any quadratic equation , we can see that there are as many solutions as it's power is. But we say either x or y is solution of this equation not both x and y are solutions of this equation. It looks the same case where OP is getting values of x but not necessarily every value is solution.