intermediate value theorem real solutions

Q: Use the Intermediate Value Theorem to prove that the equation

$$-x^3+4\sin(x)+4\cos^2(x)=0$$
has at least two solutions. You should carefully justify each of the hypothesis of the theorem.

How do I know that at least two exist because I know you're meant to look for a change in sign.

Answer

There is no problem with continuity. Hence it remains to show two intervals, where the function $f$ changes sign.
$f(0)=4$, $f(2\pi)=4-(2\pi)^3<0$. The next point needs more precise computation: $f(-\pi/2)=(\pi/2)^3-4<0$.

Hence there exist roots in $(\frac{-\pi}2,0)$ and $(0, 2\pi)$.

We used IVT in the form: If a continuous function has values of opposite sign inside an interval, then it has a root in that interval.