Quadratic equations/modulus  
Hello! 
I need help with this. I understand the solution, and everything else about the question. What I don't understand is, why does the book say that absolute value of m is greater than 1? I don't get it. Please explain it to me. Thanks
 A: The trinomial is $t^2-(m-3)t+m$, and its discriminant is $\Delta=(m-3)^2-4m$. The trinomial has at least one real root iff $\Delta\ge0$, hence $(m-3)^2-4m\ge0$. (that's not a strict inequality)
Consider the function $h(m)=(m-3)^3-4m=m^2-10m+9=(m-1)(m-9)$
$h(m)\ge0$ iff ($m\ge1$ and $m\ge9$) or ($m\le1$ and $m\le9$), that is iff $m\in ]-\infty,1]\cup[9,+\infty[$.
And you are right, $m$ could be in $[-1,1]$, which means $|m|\le1$.

It's easy to check that $|m|\le1$ would lead to solutions of the initial equation. For instance, with $m=0$, you get
$$\sin^2 t+3\sin t=\sin t(\sin t+3)$$
And this is zero for $t=k\pi$.

Let's try to finish the proof anyway.
If $m\ge9$, then the product of the roots of $f(t)=0$ is greater than $1$, and one root lies in $[-1,1]$. Hence the other does not. That means that there is a change of sign between $f(-1)$ and $f(1)$, or $f(-1)f(1)\le0$.
Then $(1-m+3+m)(1+m-3+m)\le0$, or $4(2m-2)\le0$, that is $m\le1$. Contradiction, as we supposed $m\ge9$.
So we must indeed have $m\le1$.
Now, if $m\le1$, is it certain the equation $f(t)=0$ has a solution $t\in[-1,1]$? Yes, because $f(1)f(-1)=(1-m+3+m)(1+m-3+m)=4(2m-2)\le0$, and the intermediate value theorem tells us $f(t)=0$ must have a root in $[-1,1]$.
Note that $m=1$ is also a valid value, so the inequality is not $m<1$ but $m\le1$. Then $f(t)=t^2+2t+1=(t+1)^2$, which is zero for $t=-1$. And $\sin x=-1$ for $x=-\frac\pi2+2k\pi$.
A:   
Solved it finally! I still don't fully understand why |m| > 1, but this is how I solved it. Is it correct? 
