I know that $2\cos^2 x +\sin x-1$ can't be factored, but I don't really know how to explain it. The best explanation I can come up with is "it would be like factoring $2x^2 + y -1$; it just doesn't work", but that hardly explains exactly why it wouldn't work. Does anyone know a better way of putting it?


I'm not asking if it's impossible to factor at all, I know the equation can be changed to only having sines and from there factored like a regular quadratic. That's not the question. The question is how do I put into words that the equation, exactly as written above, cannot be factored unless you change the equation into $-2 \sin^2⁡x+\sin⁡x+1$ and from there factor it as $(-2 \sin⁡x-1)(\sin⁡x-1)=0$. I've already done that latter part, I just need a way to explain why I have to do it, why it's necessary in the first place to change $2\cos^2 x +\sin x-1$ in order to factor it. I'm sorry for being unclear.


To clear up some confusion, here's what the assignment asks for:

Given the equation $2 \cos^2⁡x+\sin⁡x-1=0$,

  1. Explain why this equation cannot be factored.

  2. Use a trigonometric identity to change the equation into one that can be factored.

  3. Factor the equation.

  4. Determine all solutions in the interval 0≤x≤2π.

I've had no problems with steps 2, 3 and 4, so I did those first. However, I'm not sure how to word the answer to 1 right, because I can't think of anything else than "it just doesn't work".

  • 2
    $\begingroup$ You could use $\cos^2=1-\sin^2$ and (try to) factor the resulting polynomial. $\endgroup$
    – Ragnar
    Dec 26 '13 at 20:29
  • 3
    $\begingroup$ the resulting polynomial IS factorable.It factors to $(2\sin x +1)(1-\sin x)$ $\endgroup$
    – K. Rmth
    Dec 26 '13 at 20:30
  • $\begingroup$ It is not quite like factoring $2x^2 + y - 1$. There are lots of relationships between $\cos \theta$ and $\sin \theta$ (such as $\cos^2\theta + \sin^2\theta = 1$), while $x$ and $y$ are two independent variables. $\endgroup$ Dec 26 '13 at 20:41
  • 2
    $\begingroup$ The expression is not a polynomial, so factorization means nothing per se, but only if you tell precisely what you mean. $\endgroup$
    – egreg
    Dec 26 '13 at 21:21
  • $\begingroup$ @user107827 : who told you it couldn't be factored? $\endgroup$ Dec 27 '13 at 4:16

you know that $\cos x^2+\sin x^2=1$. Because of this, the polynomial can be written as $$ 2(1-\sin^2 x)+\sin x-1=-2\sin ^2 x+\sin x +1=(2\sin x +1)(-\sin x+1) $$

When you do not want to use any relationships between $\sin $ and $\cos$, you have two independent variables indeed, so $2x^2+y-1$. This can't be factored in general. To show that, you can try to find some values of $x$ and $y$ (do you want to 'assume' $\sin$ and $\cos $ have range $[-1,1]$) for which the expression is prime. For example $x=1$ and $y=1$.

A number $p$ is prime if its only factors (positive divisors) are $1$ and $p$ itself. $2$ is prime, but $6=2\cdot 3$ is not for example. When $2x^2+y-1$ is prime, it cannot have any factors except for $1$ and itself.
I now realize this is not a conclusive proof, because it can occur that a factor is $1$ sometimes. (See some comments on this post)

The question only says 'explain', so a full proof doesn't seem to be necessary. The only term that can be factored is the $2\cos^2x$. A factorization will look like $(\cos x+a)(2\cos x+b)$ (It is possible that the two factors in front of the $\cos$ are both $\sqrt 2$ etc, but that won't work. Lets assume they have to be integers.) Now, $2a+b=0$ and $ab=\sin x-1$. It follows that $b=-2a$, thus $-2a^2=\sin x-1$. This results in the (not so nice) factorization $$ 2\cos^2x+\sin x-1=2\left(\cos x + \sqrt{\frac{1-\sin x}2}\right)\left(\cos x - \sqrt{\frac{1-\sin x}2}\right) $$

  • $\begingroup$ I'm sorry for being unclear, but this wasn't what I was asking. If you look back at the original post, you'll see what I meant. $\endgroup$
    – Threethumb
    Dec 26 '13 at 21:09
  • $\begingroup$ I don't quite understand what you mean by finding values of x and y for which the expression is prime. I'm not natively English, so a lot of math terminology is unknown to me. What does it mean that an expression is prime? Also, if $x = 1$ and $y = 1$, then $2x^2 + y - 1 = 2$, not $0$. I'm not sure I follow what you mean. $\endgroup$
    – Threethumb
    Dec 26 '13 at 23:04
  • $\begingroup$ This answer is the right idea, but the details are wrong. One should say it like this: if one could factor the original expression, then one could substitute $\sin$ and $\cos$ with $y$ and $z$ (let's not use $x$ twice...) to arrive at a factorization of $2y^2+z-1$. Then it remains to be shown that this polynomial is in fact irreducible, which could be done either with brute force (write it as the product of two linear polynomials in general form) or with a skilled application of Eisenstein's criterion. $\endgroup$
    – Slade
    Dec 27 '13 at 0:12
  • $\begingroup$ But wouldn't writing it as the product of two linear polynomials in general form be the same as factoring it? Or am I getting you wrong? Also, I looked into Eisenstein's criterion but I didn't quite understand how to apply it in this scenario. I updated the OP to show exactly how the question goes, by the way. $\endgroup$
    – Threethumb
    Dec 27 '13 at 9:36
  • $\begingroup$ Okay, I see. That makes sense. Thanks! $\endgroup$
    – Threethumb
    Dec 27 '13 at 15:56

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