1. **Stating the problem:** We want to analyze the function $$f(x) = \frac{\cos\left(\frac{2x}{3}\right) + \sin\left(\frac{3x}{2}\right)}{2 - \cos\left(\frac{x}{4}\right)}.$$\n\n2. **Examine the numerator:** The numerator is $$\cos\left(\frac{2x}{3}\right) + \sin\left(\frac{3x}{2}\right).$$ This is a sum of cosine and sine functions with different arguments.\n\n3. **Examine the denominator:** The denominator is $$2 - \cos\left(\frac{x}{4}\right).$$ Since $\cos(\theta)$ ranges from $-1$ to $1$, the denominator ranges between $$2 - 1 = 1$$ and $$2 - (-1) = 3,$$ so it never equals zero, ensuring $f(x)$ is defined for all real $x$.\n\n4. **Simplify if possible:** There are no immediate algebraic simplifications to combine numerator or denominator further. Each function inside is in simplest trigonometric form.\n\n5. **Interpretation:** This function is a ratio of oscillating trigonometric sums.\n- Numerator oscillates between $$-\left|\cos \frac{2x}{3}\right| - \left|\sin \frac{3x}{2}\right|$$ and $$\left|\cos \frac{2x}{3}\right| + \left|\sin \frac{3x}{2}\right|.$$\n- Denominator oscillates between 1 and 3.\n\n6. **Final function:** $$f(x) = \frac{\cos\left(\frac{2x}{3}\right) + \sin\left(\frac{3x}{2}\right)}{2 - \cos\left(\frac{x}{4}\right)}.$$$\n\nThe function is defined and continuous for all real $x$ because the denominator is never zero.