Express 4cosx-3sinx in the form Rcos (x + a) where R > 0 and 0° < a < 90°. Hence, solve (4 cos x- 3 sin x)² = 4 for 0° ≤ x ≤ 360°.

$$R\cos(x + \alpha) = R(\cos x \cos \alpha - \sin x \sin \alpha)$$ $$R\cos(x + \alpha) = (R\cos \alpha)\cos x - (R\sin \alpha)\sin x$$

Comparing this to $4\cos x - 3\sin x$, we equate the coefficients:

$$R\cos \alpha = 4 \quad (1)$$ $$R\sin \alpha = 3 \quad (2)$$

Square and add equations (1) and (2):

$$(R\cos \alpha)^2 + (R\sin \alpha)^2 = 4^2 + 3^2$$ $$R^2(\cos^2 \alpha + \sin^2 \alpha) = 16 + 9$$ $$R^2(1) = 25$$ $$R = \sqrt{25} = 5 \quad (\text{since } R > 0)$$

Divide equation (2) by equation (1):

$$\frac{R\sin \alpha}{R\cos \alpha} = \frac{3}{4}$$ $$\tan \alpha = \frac{3}{4}$$ $$\alpha = \arctan\left(\frac{3}{4}\right) \approx 36.87^\circ$$

Since $0^\circ < \alpha < 90^\circ$, this value is valid.
Therefore, $4\cos x - 3\sin x = \boxed{5\cos(x + 36.87^\circ)}$.

Hence, solve $(4\cos x - 3\sin x)^2 = 4$ for $0^\circ \le x \le 360^\circ$.

$$(5\cos(x + 36.87^\circ))^2 = 4$$ $$25\cos^2(x + 36.87^\circ) = 4$$ $$\cos^2(x + 36.87^\circ) = \frac{4}{25}$$

$$\cos(x + 36.87^\circ) = \pm\sqrt{\frac{4}{25}}$$ $$\cos(x + 36.87^\circ) = \pm\frac{2}{5}$$ $$\cos(x + 36.87^\circ) = \pm 0.4$$

Let $Y = x + 36.87^\circ$. The range for $x$ is $0^\circ \le x \le 360^\circ$.
So, the range for $Y$ is $0^\circ + 36.87^\circ \le Y \le 360^\circ + 36.87^\circ$, which is $36.87^\circ \le Y \le 396.87^\circ$.

The principal value is $Y_1 = \arccos(0.4) \approx 66.42^\circ$.
Since cosine is positive in the 1st and 4th quadrants:

$$Y_1 = 66.42^\circ$$ $$Y_2 = 360^\circ - 66.42^\circ = 293.58^\circ$$

(Adding $360^\circ$ to $Y_1$ would give $426.42^\circ$, which is outside the range for $Y$).

The principal value (in the 2nd quadrant) is $Y_3 = \arccos(-0.4) \approx 113.58^\circ$.
Since cosine is negative in the 2nd and 3rd quadrants:

$$Y_3 = 113.58^\circ$$ $$Y_4 = 360^\circ - 113.58^\circ = 246.42^\circ$$

(Adding $360^\circ$ to $Y_3$ would give $473.58^\circ$, which is outside the range for $Y$).

$$x_1 = 66.42^\circ - 36.87^\circ = 29.55^\circ$$ $$x_2 = 113.58^\circ - 36.87^\circ = 76.71^\circ$$ $$x_3 = 246.42^\circ - 36.87^\circ = 209.55^\circ$$ $$x_4 = 293.58^\circ - 36.87^\circ = 256.71^\circ$$

All these values are within the range $0^\circ \le x \le 360^\circ$.

The solutions for $x$ are $\boxed{29.55^\circ, 76.71^\circ, 209.55^\circ, 256.71^\circ}$.