This is a complex analysis puzzle that seems tricky.
How can one show that for $a_k \in \mathbb{R}$, $\sum_{k=10}^{50} a_k \cos(k \theta)$ has at least four zeroes on $[0,2\pi]$? A hint is to consider the function $g(z) = \sum_{k=10}^{50} a_k z^k$ on the unit disk, whose real part on the unit circle coincides with the sum above.
The bounds $10 \to 50$ are probably overkill. What's the general trick to analyze this kind of questions?
ANSWERED IN COMMENTS: The point is that the winding number of the function around the unit circle is at least $20\pi$. Given a curve surrounding a domain $D$, the total change in the argument of any function $f(z)$ around equals (number of zeroes in $D$) - (number of poles in $D$). For us, our $g(z)$ has an order 10 zero at the origin and no poles for this polynomial, the winding number will be at least 10, which means that the real part of $g(z)$ hits zero at least 20 times.