$\text{If }\cos \mathrm{\alpha }+\cos \mathrm{\beta }=3/2\text{ and }\sin \mathrm{\alpha }+\sin \mathrm{\beta }=1/2\text{ and }\mathrm{\theta }$ is the arithmetic mean of $\mathrm{\alpha } \mathrm{and} \mathrm{\beta }$, then $\sin 2\mathrm{\theta }+\cos 2\mathrm{\theta }=$

$\begin{array}{l}\cos \mathrm{\alpha }+\cos \mathrm{\beta }=\frac{3}{2}\text{ and }\sin \mathrm{\alpha }+\sin \mathrm{\beta }=\frac{1}{2}\\ \Rightarrow 2\cos \left(\frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\right)\cos \left(\frac{\mathrm{\alpha }-\mathrm{\beta }}{2}\right)=\frac{3}{2}\text{ and }2\sin \left(\frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\right)\cos \left(\frac{\mathrm{\alpha }-\mathrm{\beta }}{2}\right)=\frac{1}{2}\Rightarrow \tan \left(\frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\right)=\frac{1}{3}\\ \because \mathrm{\theta }=\frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\Rightarrow \tan \mathrm{\theta }=\frac{1}{3}\\ \therefore \sin 2\mathrm{\theta }+\cos 2\mathrm{\theta }=\frac{2\tan \mathrm{\theta }}{1+{\tan }^2\mathrm{\theta }}+\frac{1-{\tan }^2\mathrm{\theta }}{1+{\tan }^2\mathrm{\theta }}=\frac{2(1/3)}{1+(1/3{)}^2}+\frac{1-(1/3{)}^2}{1+(1/3{)}^2}\\ =\frac{2}{3}\times \frac{9}{10}+\frac{8}{9}\times \frac{9}{10}=\frac{6}{10}+\frac{8}{10}=\frac{7}{5}.\end{array}$