The tricky thing is to pick the \(θ^μ\) without assuming that \(H\) is non-zero. The simplest way to do that is to assume that none of the coefficients of \(H\) vanish, and, since we have four unknowns (not counting \(ε_0\) and \(ε_1\)) and three equations, to set \(B = 1\). Then the first equation gives \(A = (ε_0 + H)/2\), the second equation gives \(C = D(H - A)\), and plugging everything into the third equation gives \(D^2 = -ε_1 / ε_0\), which implies that \(ε_1 = -ε_0\) and \(D = ±1\). Set \(ε_0 = -1\) to make the frame have a Lorentzian signature \(({-} \; {+} \; {+} \; {+})\), and let \(D = ε\).