Express 4 cos θ – 3 sin θ in the form R sin(θ – α), where R > 0 and 0° < α < 360°, giving α to the nearest degree. Solve the equation 4 cos θ – 3 sin θ = 3, for 0° ≤ θ ≤ 360°.

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

$$(R \cos \alpha)^2 + (R \sin \alpha)^2 = (-3)^2 + (-4)^2$$ $$R^2 (\cos^2 \alpha + \sin^2 \alpha) = 9 + 16$$ $$R^2 (1) = 25$$

Since $R > 0$,

$$R = \sqrt{25} = 5$$

Divide equation (2) by equation (1):

$$\frac{R \sin \alpha}{R \cos \alpha} = \frac{-4}{-3}$$ $$\tan \alpha = \frac{4}{3}$$

Since $R \cos \alpha = -3$ (negative) and $R \sin \alpha = -4$ (negative), $\alpha$ must be in the third quadrant.
The reference angle is $\arctan\left(\frac{4}{3}\right) \approx 53.13^\circ$.
Therefore, $\alpha = 180^\circ + 53.13^\circ = 233.13^\circ$.
Rounding to the nearest degree, $\alpha = 233^\circ$.

$$4 \cos \theta - 3 \sin \theta = 5 \sin(\theta - 233^\circ)$$

The final answer is $\boxed{5 \sin(\theta - 233^\circ)}$.

Determine the range for $\phi$.
Given $0^\circ \le \theta \le 360^\circ$:

$$0^\circ - 233^\circ \le \theta - 233^\circ \le 360^\circ - 233^\circ$$ $$-233^\circ \le \phi \le 127^\circ$$

$$\phi_1 = \arcsin(0.6) \approx 36.87^\circ$$

This value is within the range $[-233^\circ, 127^\circ]$.

Since $\sin \phi$ is positive, $\phi$ can be in the first or second quadrant.
The second general solution is $180^\circ - \phi_1$.

$$\phi_2 = 180^\circ - 36.87^\circ = 143.13^\circ$$

This value is not within the range $[-233^\circ, 127^\circ]$ as $143.13^\circ > 127^\circ$.

We also need to consider solutions by subtracting $360^\circ$.
For $\phi_1$: $36.87^\circ - 360^\circ = -323.13^\circ$. This is not within the range as $-323.13^\circ < -233^\circ$.
For $\phi_2$: $143.13^\circ - 360^\circ = -216.87^\circ$. This value is within the range $[-233^\circ, 127^\circ]$.

So, the valid values for $\phi$ are approximately $36.87^\circ$ and $-216.87^\circ$.

For $\phi = 36.87^\circ$:

$$\theta_1 = 36.87^\circ + 233^\circ = 269.87^\circ$$

For $\phi = -216.87^\circ$:

$$\theta_2 = -216.87^\circ + 233^\circ = 16.13^\circ$$

Rounding to one decimal place:

$$\theta_1 \approx 269.9^\circ$$ $$\theta_2 \approx 16.1^\circ$$

The final answers are $\boxed{\theta = 16.1^\circ, 269.9^\circ}$.