Is there any value for not defined?

Question

\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 $-1$?

Answer 1

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.

Answer 2

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!

Answer 3

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.