You have provided a time-independent wavefunction, where the wavefunction is only a function of some dimension #x#:

#\Psi(x) = Asin(kx)#

The time independent Schrödinger equation, assuming a single nonrelativistic particle is:

#\hat(H)\Psi(\vecr) = E\Psi(\vecr)#

Of course this equation states that the time-independent wavefunction is an eigenfunction of the Hamiltonian corresponding to an eigenvalue equal to the energy.

We can write the TI-equation expanded as:

#[(-ħ^2)/(2m)\grad^2+V(\vecr)]\Psi(\vecr)=E\Psi(\vecr)#

Since we are only dealing with a #\vec r# that has a component in #x#, we can use #x# explicitly:

#[(-ħ^2)/(2m)(d^2)/dx^2+V(x)]\Psi(x)=E\Psi(x)#

Our question asks us to consider the case where #V(x) \equiv V#, independent of any position. Let's also substitute in our wavefunction #\Psi(x)#:

#[(-ħ^2)/(2m)(d^2)/dx^2+V]Asin(kx)=EAsin(kx)#

#(-ħ^2)/(2m)(d^2)/dx^2(Asin(kx))+VAsin(kx)=EAsin(kx)#

#(-ħ^2)/(2m)(d)/dx(kAcos(kx))+VAsin(kx)=EAsin(kx)#

#(-ħ^2)/(2m)(-k^2Asin(kx))+VAsin(kx)=EAsin(kx)#

#(ħ^2k^2)/(2m)Asin(kx)+VAsin(kx)=EAsin(kx)#

#(ħ^2k^2)/(2m)+V=E#

Since we have found that the energy associated with this wavefunction is constant and not dependent on position, we have shown the wavefunction #\Psi(x)# satisfies the time-independent Schrödinger equation when #V# and #E# are independent of #x#.

It may also be useful to see that #ħk# is simply momentum #p#. This means we have found:

#E = (p^2)/(2m)+V#

This makes sense, since energy #E# is a sum of kinetic energy #E_k# and potential energy #E_p#. In this case, we have the classic form of kinetic energy #E_k = (p^2)/(2m)# and our potential energy is simply a constant #E_p = V#. This means our solution is valid as it obeys conservation of energy.