While solving a physics problem , I came across a equation that looks like this, $$\int_0^\pi \sigma(\theta)\,\sin(2\theta)\,d\theta=\frac{q}{\pi a^2} $$ Is there a way to solve for $\sigma(\theta)$. Understand that I am absolutely not familiar with these kinds of equations so maybe I am in the wrong track but I would still like to know it there is a unique or a family of solutions to this equation. q and $a^2$ are constants.
Edit: There is one more constraint, $$\int_0^\pi \sigma(\theta) sin(\theta)d\theta = \frac{q'}{2\pi r^2}$$ q' and r are also constants.
Since we do not know anything about the function we are searching for, we will start by imposing some things that will facilitate the search process for us.
$$I = \int ^{\pi} _{0}\sigma(\theta)\,\sin(2\theta)\,d\theta=\frac{-1}{2}(\sigma(\pi)-\sigma(0))-\frac{1}{4}\int^{\pi}_{0}\sigma^{(2)}(\theta)\sin(2\theta)d\theta$$ Therfore: $$I=\frac{-1}{2}(\sigma(\pi)-\sigma(0))+\sum^n_{k=2}\frac{(-1)^k}{2^{k+1}}(\sigma^{(2k-2)}(\pi)-\sigma^{(2k-2)}(0)) +\frac{(-1)^n}{2^{n+1}}\int ^{\pi}_{0}\sigma^{(2n)}(\theta)d\theta$$ we have $\sigma(\theta) $ polynomial of degree n+1 implies that $\sigma^{(n+2)}(\theta) =0\implies \sigma^{(2n)}(\theta) =0 ,$
This implies that : $$I =\frac{-1}{2}(\sigma(\pi)-\sigma(0))+\sum^{\lfloor \frac{n+4}{4}\rfloor}_{k=2}\frac{(-1)^k}{2^{k+1}}(\sigma^{(2k-2)}(\pi)-\sigma^{(2k-2)}(0)) =\frac{q}{\pi a^2}$$
Let $n=0$ (I'm going to look for a polynomial of degree 1)
Therfore: $$I = \frac{-1}{2}(\sigma(\pi)-\sigma(0))=\frac{q}{\pi a^2}$$ This implies that : $$\sigma(\pi)=\sigma(0)+\frac{-2q}{\pi a^2}$$
Now find the function that satisfies these conditions $\sigma(\pi)=\sigma(0)+\frac{-2q}{\pi a^2}$. It is usually assumed that: $\sigma(0)=0$
Therfore : $$\sigma(\theta)=\begin{cases} \sigma(0)=0 \\ \sigma(\pi)=\frac{-2q}{\pi a^2}\end{cases}$$
$$\boxed{\sigma(\theta)=\frac{-2q}{\pi a^2}\theta}$$