pt $<=> sin(x+\frac{\pi}{4})=\frac{m}{2}$$0\leqslant x \leqslant \frac{\pi}{2}=>\frac{\pi}{4} < x+ \frac{\pi}{4} \leqslant \frac{3\pi}{4}=> \frac{\sqrt{2}}{2} < sin(x+ \frac{\pi}{4}) \leqslant 1 => \frac{\sqrt{2}}{2}<\frac{m}{2} \leqslant 1<=> \sqrt{2}<m \leqslant2$
pt $<=> sin(x+\frac{\pi}{4})=\frac{m}{
\sqrt{2
}}$$0\leqslant x \leqslant \frac{\pi}{2}=>\frac{\pi}{4}
\leqsl
ant x+ \frac{\pi}{4} \leqslant \frac{3\pi}{4}=> \frac{\sqrt{2}}{2}
\l
eqslant sin(x+ \frac{\pi}{4}) \leqslant 1 => \frac{\sqrt{2}}{2}
\leqsl
ant\frac{m}{
\sqrt{2
}} \leqslant 1<=>
1\
leq
sl
ant
m \s
qrt
{2
}$