= Solution
For a boost of speed $v$ in the $x$ direction, with
$$
\gamma=\frac1{\sqrt{1-v^2/c^2}},
$$
the <Lorentz transformation of electromagnetic fields> is
$$
\begin{array}{lll}
E'_x=E_x,&E'_y=\gamma(E_y-vB_z),&E'_z=\gamma(E_z+vB_y),\\
B'_x=B_x,&B'_y=\gamma(B_y+vE_z/c^2),&B'_z=\gamma(B_z-vE_y/c^2).
\end{array}
$$
Expanding $E'_yB'_y+E'_zB'_z$, the mixed terms cancel and the remaining transverse terms acquire the factor $\gamma^2(1-v^2/c^2)=1$. Together with the unchanged longitudinal product, this proves the Lorentz invariant
$$
\boxed{\mathbf E'\cdot\mathbf B'=\mathbf E\cdot\mathbf B}.
$$
An independent invariant is
$$
\boxed{E^2-c^2B^2}.
$$
If only $E_y=E$ and $B_y=B$ are nonzero, then
$$
\mathbf E'=(0,\gamma E,\gamma vB),
\qquad
\mathbf B'=(0,\gamma B,-\gamma vE/c^2),
$$
and direct subtraction gives
$$
E'^2-c^2B'^2
=\gamma^2(1-v^2/c^2)(E^2-c^2B^2)
=E^2-c^2B^2.
$$
Now put $cB=\lambda E$ and $\beta=v/c$. Invariance of the dot product gives $\mathbf E'\cdot\mathbf B'=\lambda E^2/c$, while the displayed components give
$$
E'^2=\gamma^2E^2(1+\lambda^2\beta^2),
\qquad
B'^2=\frac{\gamma^2E^2}{c^2}(\lambda^2+\beta^2).
$$
Hence
$$
\boxed{\cos\theta=
\frac{\lambda(1-\beta^2)}{sqrt{(1+\lambda^2\beta^2)(\lambda^2+\beta^2)}}}.
$$
At $\beta=0$ this equals $\operatorname{sgn}\lambda$, and as $|\beta|\uparrow1$ it tends continuously to zero with the same sign. Thus boosts realize every $0\leq\theta<\pi/2$ when $\lambda>0$, and every $\pi/2<\theta\leq\pi$ when $\lambda<0$.
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