Particle in infinite well but not centred at origin

When the box is centered at origin, we get the wavefunction $$$$\psi = \sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})$$$$ But what will be the wavefunction of the particle is not centered at origin? If the box extends from $$x = x_0$$ to $$x = x_0+l$$. To solve this, I used the general solution which is $$$$\psi = A\sin(\frac{\sqrt{2mE}x}{\hbar})+B\cos(\frac{\sqrt{2mE}x}{\hbar})$$$$ then applied boundary conditions which gave me $$$$\frac{A}{B} = \tan(\frac{\sqrt{2mE}(l+2x_0)}{\hbar})$$$$ Now I'm stuck here. I don't know how to solve further. Any help will be appreciated

You can trivially obtain the solutions to the shifted problem by replacing $$x$$ with $$x-x_0$$ in the original solutions. From there, it's a matter of elementary trigonometry (i.e. the angle addition formula) to express the result in terms of $$\sin([\ldots]x)$$ and $$\cos([\ldots]x)$$.
$$\psi = A\sin([...]x+a)$$
where $$a$$ is a constant. You can easily put the boundary conditions you like.