Showing that two distinct solutions exist

Question

Here's the question 6(b) from the following image I've been trying to solve:

Let $f(x)=a_1 + a_2 \cos x + a_3 \cos 2x $. If $\left| a_1 \right| + \left| a_2 \right| < a_3 $, show that it has at least two roots in $[0, \pi ]$.

I was planning to use IVT. But it seems that

$f(0)=a_1 + a_2 + a_3 > 0$

$f\left( \frac{\pi }{2} \right) = a_1 + a_3 > |a_2| \ge 0$

$f(\pi )=a_1 - a_2 + a_3 > 0$

Had the signs been alternating, I could have used IVT. Am I going wrong somewhere?

Here's the image:

Answer 1

A hint:

Put $\cos x=:u$. Then you have to prove the following: If $|a_1|+|a_2|<a_3$ then $$q(u):=a_1+a_2 u+ a_3(2u^2-1)=0$$ has at least two solutions in the interval $[{-1},1]$.