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.