Our problem was this: describe the motion of the particle in each of the 3 dimensions. The particle starts at the origin at rest. $\overrightarrow E$ and $ \overrightarrow B$ are both constant and point in the $x$ direction.
What I don't understand is where the complexity is. The particle, when released from rest, will begin to move in the $x$ direction. It will accelerate up to a constant velocity and then stay at that velocity.
However, my classmates were deriving answers that showed that the motion of the particle in the $y$ and $z$ directions were something involving sine and cosine.
To clarify: what type of initial motion in the $yz$ direction must the particle have in order to create circular motion around the $x$-axis? Linear, quadratic, etc...?
For a charged particle in a uniform magnetic field to experience uniform circular motion, it is necessary that the magnetic field $B$ be oriented orthogonally to the particle's velocity. If what you say is true that the electric field $E$ and the magnetic field $B$ are pointing in the same direction, then there will be no uniform circular motion since, as you note, the particle will accelerate in the $x$-direction. By the Lorentz force law, the force on the particle is $$ F = qE + qv\times B = qE, $$ since $v\times B = 0$ because $v$ and $B$ are parallel. Now, if $B$ pointed in the $y$-direction, say, you may see uniform circular motion.
Let's say that the particle starts with some velocity in the $y,z$-plane. That is, $v_0 = (0, v_y, v_z)$ and $r_0 = (0, -R, 0)$. Then, the force exerted by the magnetic field on the charged particle is $F = ma = qv\times B = qvB$. Since the velocity in the $y,z$ plane is orthogonal to the magnetic field which points in the $x$-direction, the force is centripetal. And thus, we have that $a = \omega^2 R$, where $R$ is the initial radius the charge starts at with respect to the origin.
Hence, $m\omega^2R = qvB \implies \omega = \sqrt{qvB/mR}$, and it should be apparent that the magnetic field can only produce a centripetal acceleration. Hence, $$ ma = qE + qv\times B \\ m\ddot x = qE_x \\ m\omega^2R = qv_{y,z}B $$ together imply that $r(t) = \tfrac{q}{m}(\tfrac{1}{2}E_xt^2, -R\cos{\omega t}, R\sin{\omega t}).$ Working out the details is a matter of integrating a few times and understanding the parametrization of circular motion.