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 P: 675 Why is region 3 described by only $D\sin(k_1 x)$? You would want $\Psi(2a+2b) = 0$, which doesn't necessarily throw out the cosine term. In fact it might be easier to just change the coordinate system so that the origin is in the center of the well. Then due to the symmetry of the potential, you just look for symmetric and antisymmetric wavefunctions.