Let the vectors u = x i + ^2 x j + k and v = 2 ^2 x i - j - 2 x k...

For the vectors $\vec{u}$ and $\vec{v}$ to be orthogonal, their dot product must be zero.

$\vec{u} \cdot \vec{v} = (\sin x)(2\sin^2 x) + (\cos^2 x)(-1) + (1)(-2\sin x) = 0$

$\Rightarrow 2\sin^3 x - \cos^2 x - 2\sin x = 0$

Substituting $\cos^2 x = 1 - \sin^2 x$, we get:

$\Rightarrow 2\sin^3 x - (1 - \sin^2 x) - 2\sin x = 0$

$\Rightarrow 2\sin^3 x + \sin^2 x - 2\sin x - 1 = 0$

Factoring by grouping:

$\Rightarrow \sin^2 x (2\sin x + 1) - 1(2\sin x + 1) = 0$

$\Rightarrow (\sin^2 x - 1)(2\sin x + 1) = 0$

This gives $\sin^2 x = 1$ or $\sin x = -\dfrac{1}{2}$.

Case 1: $\sin^2 x = 1 \Rightarrow \sin x = 1$ or $\sin x = -1$
In the interval $[0, 4\pi]$, $\sin x = 1$ gives $x = \dfrac{\pi}{2}, \dfrac{5\pi}{2}$ (2 solutions).
$\sin x = -1$ gives $x = \dfrac{3\pi}{2}, \dfrac{7\pi}{2}$ (2 solutions).

Case 2: $\sin x = -\dfrac{1}{2}$
In the interval $[0, 4\pi]$, the solutions are $x = \dfrac{7\pi}{6}, \dfrac{11\pi}{6}, \dfrac{19\pi}{6}, \dfrac{23\pi}{6}$ (4 solutions).

Total number of solutions = $2 + 2 + 4 = 8$.

Answer: $8$