Given, $\cos \theta =-\frac{1}{\sqrt{2}}=-\cos \frac{\mathrm{\pi }}{4}$$=\cos \left(\mathrm{\pi }-\frac{\mathrm{\pi }}{4}\right) or \cos \left(\mathrm{\pi }+\frac{\mathrm{\pi }}{4}\right)$(because $\cos \theta$ is negative in second and third quadrant only)

$\therefore \theta =\frac{3\mathrm{\pi }}{4},\frac{5\mathrm{\pi }}{4}$

$\tan \theta =1=\tan \frac{\mathrm{\pi }}{4}=\tan \left(\mathrm{\pi }+\frac{\mathrm{\pi }}{4}\right)$ (because $\tan \theta$ is positive in first and third quadrant only)

$\therefore \theta =\frac{\mathrm{\pi }}{4},\frac{5\mathrm{\pi }}{4}$

Therefore, common value of $\theta$ which satisfies both the equation is $\raisebox{1ex}{$5\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$.

Therefore, general value $=2\mathrm{n\pi }+\frac{5\mathrm{\pi }}{4}=\left(2\mathrm{n}+1\right)\mathrm{\pi }+\frac{\mathrm{\pi }}{4}, \mathrm{for} \mathrm{n}=0,1,2.....$