1. **Problem Statement:** Find the complex exponential Fourier series representation of the signal $$m(\beta) = \frac{\sin 6\beta}{2} + \frac{\cos 2\beta}{3}$$. 2. **Recall the complex exponential Fourier series formula:** $$m(\beta) = \sum_{n=-\infty}^{\infty} c_n e^{jn\beta}$$ where $$c_n = \frac{1}{2\pi} \int_{-\pi}^{\pi} m(\beta) e^{-jn\beta} d\beta$$. 3. **Express sine and cosine in terms of complex exponentials:** $$\sin 6\beta = \frac{e^{j6\beta} - e^{-j6\beta}}{2j}$$ $$\cos 2\beta = \frac{e^{j2\beta} + e^{-j2\beta}}{2}$$ 4. **Rewrite the signal using these expressions:** $$m(\beta) = \frac{1}{2} \cdot \frac{e^{j6\beta} - e^{-j6\beta}}{2j} + \frac{1}{3} \cdot \frac{e^{j2\beta} + e^{-j2\beta}}{2} = \frac{e^{j6\beta} - e^{-j6\beta}}{4j} + \frac{e^{j2\beta} + e^{-j2\beta}}{6}$$ 5. **Identify the Fourier coefficients $$c_n$$ by matching terms:** - For $$n=6$$: $$c_6 = \frac{1}{4j}$$ - For $$n=-6$$: $$c_{-6} = -\frac{1}{4j}$$ (since $$-e^{-j6\beta}$$ term) - For $$n=2$$: $$c_2 = \frac{1}{6}$$ - For $$n=-2$$: $$c_{-2} = \frac{1}{6}$$ - For all other $$n$$, $$c_n = 0$$. 6. **Final complex exponential Fourier series representation:** $$m(\beta) = \frac{1}{4j} e^{j6\beta} - \frac{1}{4j} e^{-j6\beta} + \frac{1}{6} e^{j2\beta} + \frac{1}{6} e^{-j2\beta}$$ This matches the original signal exactly. **Summary:** The signal has nonzero Fourier coefficients only at $$n=\pm 6$$ and $$n=\pm 2$$ with values as above.