Let $g(\theta)$ be a known real-valued function with domain $[0, 2\pi]$. Given that:
$$\int_{0}^{2\pi} f(t) \sin ^2 (t-\theta) dt = g(\theta)$$
How would I solve for the unknown real-valued function $f(t)$? Is the solution unique?
We can assume $f$ is integrable. The function $g$ is known, so its Fourier series and Fourier transform can be readily computed. But how to use them to find $f$?
Looking for the Fourier series of $f$: $$ f(t) = \frac{A_0}{2}+\sum_{n=1}^\infty(A_n\cos nt+B_n\sin nt) \tag{1} $$ When this is multiplied by $$\sin^2(t-\theta ) = \frac12-\frac12\cos(2(t-\theta )) = \frac12-\frac12\cos 2\theta \cos 2t - \frac12 \sin 2\theta \sin 2t $$ and integrated over the circle, most terms cancel out due to orthogonality. You end up with $$ \frac12\left(\pi A_0 - A_2\cos 2\theta -B_2\sin 2\theta\right) \tag{2} $$ And this is what you are equating to $g$... so there is no solution unless $g$ is of the form $(2)$. And if it happens to be of that form, there are infinitely many solutions because all coefficients in $(1)$ except $A_0,A_2,B_2$ can be anything.
Summary: the problem is that the kernel $\sin^2(t-\theta)$ has few harmonics. Something like the Gaussian kernel would allow you to solve the equation for a wide class of functions $g$.