Solution. Considering the basic trigonometric identity, we get $\cos \left(\pi x^{2}\right)+\cos \left(\pi x^{2}-4 \pi x\right)=0 \quad \Rightarrow \quad 2 \cos \left(\pi x^{2}-2 \pi x\right) \cos (2 \pi x)=0$.

Therefore, $\left[\begin{array}{l}\cos \left(\pi x^{2}-2 \pi x\right)=0, \\ \cos (2 \pi x)=0,\end{array} \Rightarrow\left[\begin{array}{l}\pi x^{2}-2 \pi x=\frac{\pi}{2}+\pi k, k \in Z, \\ 2 \pi x=\frac{\pi}{2}+\pi n, n \in Z,\end{array} \Rightarrow\left[\begin{array}{l}x^{2}-2 x-\frac{1}{2}+k=0, k \in Z, \\ x=\frac{1}{4}+\frac{n}{2}, n \in Z,\end{array}\right.\right.\right.$ Solving the first equation, we get $x=1 \pm \sqrt{k+3 / 2}$, and then the sequence of positive roots of the equation will be as follows:

$0.25 ; \frac{2-\sqrt{2}}{2} ; 0.75 ; 1.25 ; \frac{\sqrt{13}-1}{2} ; 1.5 ; 1+\sqrt{1 / 2}, \ldots$

Answer: 0.75
2. Solve the equation $\cos \left(\pi x^{2}\right)-\cos ^{2}\left(\frac{\pi x^{2}}{2}\right)+1+\cos \left(\pi x^{2}-4 \pi x\right)=\sin ^{2}\left(\frac{\pi x^{2}}{2}\right)$. In your answer, specify the third term of the increasing sequence of all positive roots of the equation.