Question

Как решить систему уравнений x+y=п/2 sinx+siny=-корень из 2 я пытался решить но не понял что делать дальше

Answer 1

$\left \{ {{x+y=\frac{\pi}{2}} \atop {sin(x)+sin(y)=-\frac{\sqrt{2}}{2}}} \right. \\\\ \left \{ {{y=\frac{\pi}{2}-x} \atop {sin(x)+sin(\frac{\pi}{2}-x)=-\frac{\sqrt{2}}{2}}} \right. \\\\ \left \{ {{y=\frac{\pi}{2}-x} \atop {sin(x)+sin(\frac{\pi}{2}-x)=-\frac{\sqrt{2}}{2}}} \right. \\\\ \left \{ {{y=\frac{\pi}{2}-x} \atop {sin(x)+cos(x)=-\frac{\sqrt{2}}{2}}} \right. \\\\ \left \{ {{y=\frac{\pi}{2}-x} \atop {\sqrt{2}sin(x+\frac{\pi}{4})=-\frac{\sqrt{2}}{2}}} \right. \\\\$

$\left \{ {{y=\frac{\pi}{2}-x} \atop {sin(x+\frac{\pi}{4})=-\frac{1}{2}}} \right. \\\\ \left \{ {{y=\frac{\pi}{2}-x} \atop {x+\frac{\pi}{4}=(-1)^k*\frac{\pi}{6}+\pi k,\ k\in Z}} \right. \\\\ \left \{ {{y=\frac{\pi}{2}+\frac{\pi}{4}-(-1)^k*\frac{\pi}{6}-\pi k,\ k\in Z} \atop {x=-\frac{\pi}{4}+(-1)^k*\frac{\pi}{6}+\pi k,\ k\in Z}} \right. \\\\ \left \{ {{y=\frac{3\pi}{4}+(-1)^{k+1}*\frac{\pi}{6}+\pi k,\ k\in Z} \atop {x=-\frac{\pi}{4}+(-1)^k*\frac{\pi}{6}+\pi k,\ k\in Z}} \right. \\\\$