What is the general solution to $$f(t+2\pi)+f(t)=0 ?$$I realise it must be a combination of sines/cosines with constants and the appropriate frequencies but I would rather be sure about it.
(Motivation: see the last line in Wave Equation with One Non-Homogeneous Boundary Condition)
This is a subset of the functions that have period $4\pi$, because:
$$f(x+4\pi) = -f(x+2\pi)=f(x)$$
The functions of of period $4\pi$ are (possibly infinite) linear combinations of $\sin(nx/2)$ and $\cos(nx/2)$ for $n$ positive integers.
In turn, the subset you want are of the ones generated by $\sin(nx/2)$ and $\cos(nx/2)$ with $n$ odd.
It is easy to see that when $n$ is odd the functions satisfy this equation, and that when $n$ is even, they do not, but what about the case of complex combinations with the $n$ even? Essentially, you need to find the Fourier series for $f$ on $[0,4\pi]$ to show that the even coefficients are zero.
This is because we can show that if $g(x+2\pi)=g(x)$ and $f(x)=-f(x+2\pi)$ then $$0=\int_{0}^{4\pi} f(x)g(x)\,dx$$
Obviously, this assumes $f$ is "nice enough" to compute a Fourier series. There are non-measurable $f$ which satisfy the above equation. But if we restrict the question to the types of functions for which Fourier analysis is effective, then this will get all of them.
Any periodic function with (not necessarily least) period $4\pi$ that has the second half of it's cycle as the negative of the first half, i.e. the concatenation (to make periodic) of
$$f(x)=\begin{cases} g(x) &\text{ for } 0 \leq x< 2\pi \\ -g(x-2\pi) &\text{ for } 2\pi \leq x< 4\pi \\ \end{cases}$$
for any function $g$. Just follows exactly from the definition.
