6.1. Find the greatest root of the equation $|\sin (2 \pi x)-\cos (\pi x)|=|| \sin (2 \pi x)|-| \cos (\pi x) \|$, belonging to the interval $\left(\frac{1}{4} ; 2\right)$.
Answer: 1.5. Solution. Since $|A-B|=|| A|-| B|| \Leftrightarrow A^{2}-2 A B+B^{2}=A^{2}-2|A B|+B^{2} \Leftrightarrow$ $A B=|A B| \Leftrightarrow A B \geq 0$, then $\sin 2 \pi x \cdot \cos \pi x \geq 0 \Leftrightarrow 2 \sin \pi x \cdot \cos ^{2} \pi x \geq 0 \Leftrightarrow\left[\begin{array}{c}\cos \pi x=0, \\ \sin \pi x \geq 0\end{array} \Leftrightarrow\right.$ $\left[\begin{array}{c}\pi x=\frac{\pi}{2}+\pi k, \\ 2 \pi n \leq \pi x \leq \pi+2 \pi n\end{array} \Leftrightarrow\left[\begin{array}{c}x=-\frac{1}{2}+2 p, \\ 2 n \leq x \leq 1+2 n .\end{array}\right.\right.$ In the interval specified in the condition, $x \in\left(\frac{1}{4} ; 1\right] \cup\left\{\frac{3}{2}\right\}$