Given, $2\theta - {\cos ^2}\theta + \sqrt 2 = 0$ $\Rightarrow 2\theta + \sqrt 2 = {\cos ^2}\theta$ $\Rightarrow 2\theta + \sqrt 2 = {{1 + \cos 2\theta } \over 2}$ $\Rightarrow 4\theta + 2\sqrt 2 = 1 + \cos 2\theta = y$ (Assume) $\therefore$ $y = 4\theta + 2\sqrt 2$ and $y = 1 + \cos 2\theta$ For $y = 1 + \cos 2\theta$ when $\theta = 0$, $y = 1 + 1 = 2$ when $\theta = {\pi \over 4}$, $y = 1 + \cos {\pi \over 2} = 1$ $\theta = {\pi \over 2}$, $y = 1 + \cos \pi = 1 - 1 = 0$ For $y = 4\theta + 2\sqrt 2$ when $\theta = 0$, $y = 2\sqrt 2$ when $\theta = {\pi \over 2}$, $y = 2\pi + 2\sqrt 2$ $= 2(\pi + \sqrt 2 )$ $= 2(3.14 + 1.41)$ $= 2(4.55)$ $= 9.1$ when $\theta = - {\pi \over 2}$, $y = - 2\pi + 2\sqrt 2$ $= 2( - \pi + \sqrt 2 )$ $= 2( - 3.14 + 1.41)$ $= - 3.46$ $\therefore$ Two graph cut's at only one point so one solution possible.