Question

Помогите, пожалуйста

Answer 1

$\left \{ {Sinx+Siny=1} \atop {Cosx-Cosy= \sqrt{3} }} \right.\\\\ :\left \{ {{2Sin \frac{x+y}{2}Cos \frac{x-y}{2}=1 } \atop {-2Sin \frac{x+y}{2}Cos \frac{x-y}{2}= \sqrt{3} }} \right. \\\\ tg \frac{x-y}{2} =- \sqrt{3}\\\\ \frac{x-y}{2} =- \frac{ \pi }{3}+ \pi n\\\\ \left \{ {{Sin \frac{x+y}{2}Cos(- \frac{ \pi }{3} + \pi n)= \frac{1}{2} } \atop {Sin \frac{x+y}{2}Sin(- \frac{ \pi }{3}+ \pi n=- \frac{ \sqrt{3} }{2} }} \right.$

$\left \{ {{Sin \frac{x+y}{2}*(-1) ^{n}Cos(- \frac{ \pi }{3})= \frac{1}{2} } \atop {Sin \frac{x+y}{2}*(-1) ^{n} Sin(- \frac{ \pi }{3})=- \frac{ \sqrt{3} }{2} }} \right.\\\\ \left \{ {{Sin \frac{x+y}{2}*(-1) ^{n}* \frac{1}{2}= \frac{1}{2} } \atop {Sin \frac{x+y}{2}*(-1) ^{n}*(- \frac{ \sqrt{3} }{2} )=- \frac{ \sqrt{3} }{2} }} \right.\\\\ \left \{ {{Sin \frac{x+y}{2}*(-1) ^{n} =1 } \atop {Sin \frac{x+y}{2}*(-1) ^{n} =1 }} \right. \\\\Sin \frac{x+y}{2} =(-1) ^{n} \\\\n =2k ; Sin \frac{x+y}{2}=1$

$\frac{x+y}{2} = \frac{ \pi }{2}+2 \pi m\\\\n=2k-1;Sin \frac{x+y}{2}=-1\\\\ \frac{x+y}{2}=- \frac{ \pi }{2}+2 \pi m\\\\ \left \{ {{ \frac{x-y}{2}=- \frac{ \pi }{3} + \pi n } \atop { \frac{x+y}{2}=(-1) ^{n} \frac{ \pi }{2}+2 \pi m }} \right.\\\\ \left \{ {{x-y=- \frac{2 \pi }{3} +2 \pi n} \atop {x+y=(-1) ^{n} \pi +4 \pi m }} \right.\\\\ \left \{ {{2x=(-1) ^{n} \pi+ 4 \pi m- \frac{2 \pi }{3} +2 \pi n} \atop {2y=(-1) ^{n} \pi +4 \pi m-(- \frac{2 \pi }{3}+2 \pi n) }} \right.$

$\left \{ {{2x=(-1) ^{n} \pi - \frac{2 \pi }{3}+2 \pi (2m+n) } \atop {2y=(-1) ^{n} \pi + \frac{2 \pi }{3}+2 \pi (2m-n) }} \right.\\\\ \left \{ {{x=(-1) ^{n} \frac{ \pi }{2}- \frac{ \pi }{3} + \pi (2m+n) } \atop {y=(-1) ^{n} \frac{ \pi }{2}+ \frac{ \pi }{3}+ \pi (2m-n) }} \right.$