If $\mathrm{\theta }$ is acute angle between two lines ${\mathrm{x}}^2+6\mathrm{xy}+{\mathrm{y}}^2=0$ then the value of $\sqrt{2}\mathrm{sin\theta }+\mathrm{cos\theta }$ is

If $\mathrm{\theta }$ is acute angle between pair of lines represented by the equation ${\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2=0$ then $\mathrm{cos\theta }=\frac{\left|\mathrm{a}+\mathrm{b}\right|}{\sqrt{{(\mathrm{a}-\mathrm{b})}^2+{\left(2\mathrm{h}\right)}^2}}$
For the equation ${\mathrm{x}}^2+6\mathrm{xy}+{\mathrm{y}}^2=0,\mathrm{a}=1,\mathrm{h}=3,\mathrm{b}=1$
Hence, 
$\begin{array}{c}\mathrm{cos\theta }=\frac{2}{6}\\ =\frac{1}{3}\end{array}$
 
Use the right triangle for $\mathrm{sin\theta }=\frac{2\sqrt{2}}{3}$
Hence, 
$\begin{array}{c}\sqrt{2}\sin \text{θ}+\cos \text{θ}=\sqrt{2}\left(\frac{2\sqrt{2}}{3}\right)+\frac{1}{3}\\ =\frac{4}{3}+\frac{1}{3}\\ =\frac{5}{3}\end{array}$
Therefore, $\sqrt{2}\sin \text{θ}+\cos \text{θ}=\boxed{\frac{5}{3}}$