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In my book, it's written that we can guess how many roots an equation might have and where they approximately are by its graph or the table of function values without trying to solve it. There's an example afterward, doing that for $f(x)=\sin x -x +0.5=0$ starting from $-1\leq \sin x \leq 1$.

The problem is, it's not always that easy to guess how many roots it has. In the exercise at the end of this section, it's asking about $\sin x=x(x-2)(x-3)$ and $x^2 \cos^2 x-1=0$. I don't know what to do about them and where should I start from.

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    @Thomas: Yes, RossMillikan has answered it. Thank you.2011-12-29

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There are some general rules, but no one specific recipe. For your first, you should think about the fact that $|\sin x | \le 1$ which would lead to the thought that there should be roots near the right side zeros: $0, 2, 3$. As $\sin 0=0$, that is a root and $\sin 3$ is very close to $0$, so the root should be close there. Then you can multiply out the right side: $\sin x=x^3-5x^2+6x$ and see that it gets large and positive as $x \to \infty$ and large and negative as $x \to -\infty$, so there won't be roots too far from the origin. At $x=3$ (the greatest zero of the right), the derivative of the right is $3$, so the right will get greater than $1$ at about $\frac{10}3$ and we can stop looking. Similarly, the derivative of the right at $0$ is $6$, so the right will be less than $-1$ by about $\frac{-1}{6}$. This Alpha graph suggests this is a reasonable guess.

For $x^2 \cos^2 x-1=0$ you can observe that $0$ is not a root and write it as $\cos^2 x=\frac1{x^2}$ or $\cos x=\pm\frac{1}{x}$. The equation is symmetric around $0$ so we will assume $x \gt 0$. There are no roots below $1$ and we can write $\frac2{x^2}=1+\cos 2x$. The left side is positive but decreasing to $0$ and the right side touches $0$ when $x=(n+\frac12 )\pi$. I would expect a pair of roots near each of these points. There is also a single one around $2$ as we get into range.

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    I meant [this one](http://www.wolframalpha.com/input/?i=plot%20x%5E2%20%5Ccos%5E2%20x-1%3D0%20from%20-1%20to%201&t=crmtb01), sorry. Apparently it has more than six roots, infinitely many roots as you said. Thank you for the great answer.2011-12-29