Question

Решить систему уравнений!!!! (фото) ​

Answer 1

$\displaystyle \left \{ {{x-y=\frac{2\pi }{3}} \atop {sinx-siny=\frac{3}{2}}} \right. \\\\x=\frac{2\pi }{3}+y\\\\sin(\frac{2\pi }{2}+y)-siny=\frac{3}{2}\\\\sin\frac{2\pi }{3}*cosy+cos\frac{2\pi }{3}*siny-siny=\frac{3}{2}\\\\\frac{\sqrt{3}}{2}cosy-\frac{1}{2}siny-siny=\frac{3}{2}$

$\displaystyle \frac{\sqrt{3}}{2}cosy-\frac{3}{2}siny=\frac{3}{2} \bigg|:\sqrt{3}\\\\\frac{1}{2}cosy-\frac{\sqrt{3}}{2}siny=\frac{\sqrt{3}}{2}\\\\cos\frac{\pi }{3}*cosy-sin\frac{\pi }{3}siny=\frac{\sqrt{3}}{2}\\\\cos(\frac{\pi }{3}+y)=\frac{\sqrt{3}}{2}$

$\displaystyle \frac{\pi }{3}+y_1=\frac{\pi }{6}+2\pi n, n \in Z;\\\\\frac{\pi }{3}+y_2=-\frac{\pi }{6}+2\pi n, n \in Z$

$\displaystyle y_1=\frac{\pi }{6}-\frac{\pi }{3}+2\pi n=-\frac{\pi }{6}+2\pi n\\\\x_1=\frac{2\pi }{3}-\frac{\pi }{6}-2\pi n=\frac{\pi }{2}-2\pi n$

$\displaystyle y_2=-\frac{\pi }{3}-\frac{\pi }{6}+2\pi n=-\frac{\pi }{2}+2\pi n\\\\x_2=\frac{2\pi }{3}-\frac{\pi }{2}-2\pi n=\frac{\pi }{6}-2\pi n$