Informally you can recover the coefficient by averaging both sides against $\sin mx$:
$$\frac{1}{\pi}\int_{-\pi}^{\pi}\left( a_0 + \sum_{n=1}^\infty a_n\sin(nx) + b_n\cos(nx)\right)\sin(mx)\, dx =\frac{1}{\pi}\int_{-\pi}^\pi 4\sin x \sin (mx)\, dx, $$
because the left hand side becomes then
$$a_0 \frac{1}{\pi}\int_{-\pi}^\pi\sin(mx)\, dx+ \sum_{n=1}^\infty a_n\frac{1}{\pi}\int_{-\pi}^\pi\sin(nx)\sin(mx)\, dx+b_n\frac{1}{\pi}\int_{-\pi}^\pi\cos(nx)\sin(mx)\, dx =a_m $$
while the left hand side is $4\delta_{1m}$. This procedure, similar to the original reasoning by Fourier, requires the interchange of integral and sum in left hand side, which can be done if the series converges uniformly. However, that request is excessively restrictive: turns out that convergence in $L^2$ sense will suffice. This can be proven by means of Hilbert space methods.
You know that
$$f(x) = \frac{a_0}{2} + \sum_{n = 1}^\infty (a_n \cos(nx) + b_n\sin(nx)), \quad x \in [-\pi,\pi].$$
So
$$\int_{-\pi}^\pi f(x)g(x)\, dx = \frac{a_0}{2}\int_{-\pi}^\pi g(x)\, dx + \sum_{n = 1}^\infty \left(a_n \int_{-\pi}^\pi \cos(nx)g(x)\, dx + b_n \int_{-\pi}^\pi \sin(nx)g(x)\, dx\right).$$
This step is justified by the uniform convergence assumptions given in the problem. By definition of the Fourier coefficients $\alpha_n$ and $\beta_n$, we have
$$\pi \alpha_0 = \int_{-\pi}^\pi g(x)\, dx,$$
$$\pi \alpha_n = \int_{-\pi}^\pi \cos(nx)g(x)\, dx,\quad n \ge 1$$
and
$$\pi \beta_n = \int_{-\pi}^\pi \sin(nx)g(x)\, dx,\quad n \ge 1.$$
Therefore
$$\int_{-\pi}^\pi f(x)g(x)\, dx = \frac{\pi a_0\alpha_0}{2} + \sum_{n = 1}^\infty (\pi a_n\alpha_n + \pi b_n \beta_n),$$
or
$$\int_{-\pi}^\pi f(x)g(x)\, dx = \frac{a_0 \alpha_0}{2} + \sum_{n = 1}^\infty (a_n \alpha_n + b_n \beta_n).$$
Best Answer
This DE does not have a periodic solution valid on $\mathbb R$, so this method fails. The actual solutions is $e^{-x}-e^{x}$