Solve the trigonometric equation 4\[2 x\] + 4 x = 5 for 0^ x 360^.

From this identity, we can write $\sin^2 x = 1 - \cos^2 x$.
Substitute this into the given equation:

$$4(1 - \cos^2 x) + 4\cos x = 5$$

$$4 - 4\cos^2 x + 4\cos x = 5$$

Move all terms to one side to set the equation to zero:

$$-4\cos^2 x + 4\cos x + 4 - 5 = 0$$ $$-4\cos^2 x + 4\cos x - 1 = 0$$

Multiply by -1 to make the leading coefficient positive:

$$4\cos^2 x - 4\cos x + 1 = 0$$

This is a perfect square trinomial, which can be factored as $(2\cos x - 1)^2 = 0$.
Alternatively, let $y = \cos x$. The equation becomes $4y^2 - 4y + 1 = 0$.
Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$$y = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(4)(1)}}{2(4)}$$ $$y = \frac{4 \pm \sqrt{16 - 16}}{8}$$ $$y = \frac{4 \pm \sqrt{0}}{8}$$ $$y = \frac{4}{8}$$ $$y = \frac{1}{2}$$

So, $\cos x = \frac{1}{2}$.

We need to find angles $x$ for which $\cos x = \frac{1}{2}$.
The reference angle for $\cos x = \frac{1}{2}$ is $60^\circ$.
Since $\cos x$ is positive, $x$ lies in the first and fourth quadrants.
In the first quadrant:

$$x = 60^\circ$$

In the fourth quadrant:

$$x = 360^\circ - 60^\circ = 300^\circ$$

Both $60^\circ$ and $300^\circ$ are within the specified range $0^\circ \le x \le 360^\circ$.

The solutions are $60^\circ, 300^\circ$.