Do there exist $a_k$ and $b_k$ so the equation $\sum\limits_{k=1}^{n} (a_k \sin(kx) + b_k \cos(kx)) = 0$ has no roots?

Question

Do there exist real numbers $a_1, a_2, ..., a_n$ and $b_1, b_2, ..., b_n$ such that the equation $$\sum\limits_{k=1}^{n} (a_k \sin(kx) + b_k \cos(kx)) = 0$$ has no solutions?

Answer 1

It is as sos440 says: $$ \int_0^{2\pi} \left[ \sum\limits_{k=1}^{n} (a_k \sin(kx) + b_k \cos(kx)) \right] \; dx = 0 $$ A strictly positive or strictly negative function cannot integrate to zero. Thus $f(x) = \sum\limits_{k=1}^{n} (a_k \sin(kx) + b_k \cos(kx))$ is neither. So there exist real numbers $a$ and $b$ with $f(a) > 0$, $f(b) < 0$. By intermediate value theorem, $f(x) = 0$ for some $x$.