Prove that a function has a root in interval, when you know, by Rolle, there is a root of the derivative, but you don't know it.

Question

Let $a_1,\ a_2,\ b_1,\ b_2 \in \mathbb{R}$. Prove that:

$$a_1\cos(x) + a_2\sin(x) + b_1\cos(2x)+b_2\sin(2x)=0$$ has at least one root in $(0,2\pi)$.

---

I started by creating a function: $f:[0,2\pi]\rightarrow\mathbb{R}$, $f(x) = a_1\cos(x) + a_2\sin(x) + b_1\cos(2x)+b_2\sin(2x)$ and computed $f(0)$ and $f(2\pi)$ then noticed that $f(0) = f(2\pi)$. So, by Rolle's Theorem, there exists such $c \in (0,2\pi)$ for which $f'(c) = 0$.

That's good. I know there exists a root for $f'(x) = 0$ but I don't know its sign as to get a relation, to be able to compare the sing of $f(0)=a_1+b_1$ with $f(c)$.

What can I do? Thank you!

Answer 1

I just realised that you can use a more straightforward method than the one I proposed in comment. All you have to do is to integrate f between $0$ and $2\pi$.

$$\int_0^{2\pi}a_1cos(x)+a_2sin(x)+b_1cos(2x)+b_2sin(2x) =[a_1sin(x)-a_2cos(x)+\frac{b_1}{2}sin(2x)-\frac{b_2}{2}cos(2x)]_0^{2\pi} =0$$

Since the integral is $0$, $f$ verify one of theses propositions.

1. $f$ is always $0$ and therefore $f(x)=0$ has an infine number of solutions.
2. $f$ has positives negatives values "cancelling" each other. Since f is continuous, from intermediate value theorem, it has at least one root in $[0,2\pi]$