= Solution
In the $S_z$ <eigenbasis>, let a candidate <eigenvector> be $(a,b)^{\mathsf T}$. The <eigenvalue equation> is $b=(2s_x/\hbar)a$ and $a=(2s_x/\hbar)b$, so a nonzero vector requires $(2s_x/\hbar)^2=1$. Choosing normalized phases gives
$$
\boxed{|+x\rangle=\frac{|+z\rangle+|-z\rangle}{\sqrt2}=\frac1{\sqrt2}\begin{pmatrix}1\\1\end{pmatrix},\quad s_x=\frac\hbar2,}
$$
$$
\boxed{|-x\rangle=\frac{|+z\rangle-|-z\rangle}{\sqrt2}=\frac1{\sqrt2}\begin{pmatrix}1\\-1\end{pmatrix},\quad s_x=-\frac\hbar2.}
$$
The two <spin eigenstates> are <orthogonal>, as required for distinct <eigenvalues> of a <self-adjoint operator>.
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