= Solution
For $E_0\ne0$, put $\beta=v/c$. Part (c) gives
$$
\begin{aligned}
E'_y&=\gamma E_0(1-\beta\sin\theta),& E'_z&=\gamma E_0\beta\cos\theta,\\
cB'_y&=\gamma E_0\cos\theta,&cB'_z&=\gamma E_0(\sin\theta-\beta).
\end{aligned}
$$
The vanishing of their cross product is equivalent to
$$
(1-\beta\sin\theta)(\sin\theta-\beta)-\beta\cos^2\theta=0,
\quad\text{or}\quad\sin\theta(1+\beta^2)-2\beta=0.
$$
For $0<\theta<\pi/2$ the two roots are $\tan(\theta/2)$ and $\cot(\theta/2)$. Only the first has $|\beta|<1$. The continuous endpoint $\theta=0$ gives $\beta=0$. Thus \b[the physical boost is]
$$
\boxed{v=c\tan(\theta/2),\qquad0\le\theta<\pi/2.}
$$
Substitution gives $\boldsymbol E'=c\boldsymbol B'$, so the fields point in the same direction. At $\theta=\pi/2$ the formal root is $v=c$, which is not an inertial observer's velocity. The nonzero perpendicular equal-strength fields then have both <electromagnetic field invariants> zero. If they were parallel in a finite frame, those invariants would force both fields to vanish, impossible under an invertible <Lorentz transformation>. Hence \b[no finite inertial boost makes them parallel in this case]. If $E_0=0$, both fields are zero in every frame and no velocity is singled out.
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