داریم: $\cos (\theta )=\sin (\frac{\pi }{2}-\theta )\Rightarrow \cos (\frac{\pi }{2}{{\cos }^{2}}x)=\sin (\frac{\pi }{2}-\frac{\pi }{2}{{\cos }^{2}}x)=\sin (\frac{\pi }{2}{{\sin }^{2}}x)$  $\Rightarrow 2\sin (\frac{\pi }{2}{{\sin }^{2}}x)=\sqrt{2}\Rightarrow \sin (\frac{\pi }{2}{{\sin }^{2}}x)=\frac{\sqrt{2}}{2}$  $\left\{ \begin{align}  & \frac{\pi }{2}{{\sin }^{2}}x=\frac{\pi }{4}\Rightarrow {{\sin }^{2}}x=\frac{1}{2}\Rightarrow \sin x=\pm \frac{\sqrt{2}}{2} \\ & \frac{\pi }{2}{{\sin }^{2}}x=\frac{3\pi }{4}\Rightarrow {{\sin }^{2}}x=\frac{3}{2}\Rightarrow \otimes (0\le {{\sin }^{2}}x\le 1) \\ \end{align} \right.\Rightarrow x=\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$ پس معادله‌ی فوق در بازه‌ی مذکور دارای چهار جواب است.