1. **State the problem:** Find the Complex Exponential Fourier Series representation of the signal $$s(t) = 4 \sin\left(2t + \frac{\pi}{6}\right)$$. 2. **Recall the formula:** The complex exponential form of a sinusoid $$A \sin(\omega t + \phi)$$ can be expressed using Euler's formula: $$\sin(\theta) = \frac{e^{j\theta} - e^{-j\theta}}{2j}$$ 3. **Apply Euler's formula:** Substitute $$\theta = 2t + \frac{\pi}{6}$$: $$s(t) = 4 \sin\left(2t + \frac{\pi}{6}\right) = 4 \cdot \frac{e^{j(2t + \frac{\pi}{6})} - e^{-j(2t + \frac{\pi}{6})}}{2j}$$ 4. **Simplify the expression:** $$s(t) = \frac{4}{2j} \left(e^{j2t} e^{j\frac{\pi}{6}} - e^{-j2t} e^{-j\frac{\pi}{6}}\right) = \frac{2}{j} \left(e^{j2t} e^{j\frac{\pi}{6}} - e^{-j2t} e^{-j\frac{\pi}{6}}\right)$$ 5. **Rewrite $$\frac{1}{j}$$:** Recall that $$\frac{1}{j} = -j$$, so: $$s(t) = 2(-j) \left(e^{j2t} e^{j\frac{\pi}{6}} - e^{-j2t} e^{-j\frac{\pi}{6}}\right) = -2j \left(e^{j2t} e^{j\frac{\pi}{6}} - e^{-j2t} e^{-j\frac{\pi}{6}}\right)$$ 6. **Express the Fourier series coefficients:** The complex exponential Fourier series representation is: $$s(t) = C_1 e^{j2t} + C_{-1} e^{-j2t}$$ where $$C_1 = -2j e^{j\frac{\pi}{6}}, \quad C_{-1} = 2j e^{-j\frac{\pi}{6}}$$ 7. **Final answer:** $$\boxed{s(t) = -2j e^{j\frac{\pi}{6}} e^{j2t} + 2j e^{-j\frac{\pi}{6}} e^{-j2t}}$$ This is the complex exponential Fourier series representation of the given sinusoidal signal.