Помогите решить систему тригонометрических уравнений.

Решите задачу:

$\left \{ {{x+y = \pi } \atop {sinx + siny = 1}} \right. \\ \\ \left \{ {{x = \pi - y } \atop {sinx + siny = 1}} \right. \\ \\ \left \{ {{x+y = \pi } \atop {sin( \pi - y) + siny = 1}} \right. \\ \\ \left \{ {{x+y = \pi } \atop {siny + siny = 1}} \right. \\ \\ \left \{ {{x+y = \pi } \atop { 2siny = 1}} \right. \\ \\ \left \{ {{x+y = \pi } \atop { siny = \frac{1}{2} }} \right. \\ \\ \left \{ {{x+y = \pi } \atop { y = \frac{ \pi }{6} }} \right.$
$\left \{ {{y= \frac{ \pi }{6} } \atop {x= \frac{5 \pi }{6} }} \right.$

$\left \{ {{x+y= \frac{ \pi }{2} } \atop {sin^2x-sin^2y = 1 }} \right. \\ \\ \left \{ {{y= \frac{ \pi }{2} -x} \atop {sin^2x - sin^2( \frac{ \pi }{2} -x) = 1}} \right. \\ \\ \left \{ {{y= \frac{ \pi }{2} -x} \atop {sin^2x - cos^2x = 1}} \right. \\ \\ \left \{ {{y= \frac{ \pi }{2} -x} \atop {-cos2x= 1}} \right. \\ \\ \left \{ {{y= \frac{ \pi }{2} -x} \atop {cos2x= -1}} \right. \\ \\ \left \{ {{y= \frac{ \pi }{2} -x} \atop {2x = - \pi }} \right.$
$\left \{ {{y= \frac{ \pi }{2} -x} \atop {x = - \frac{ \pi }{2} }} \right. \\ \\ \left \{ {{y= \pi } \atop {x = - \frac{ \pi }{2} }} \right.$