sin2x/sin(π-x)=√2
2sinxcosx/sinx=√2
sinx≠0
2cosx=√2
cosx=√2/2
x=-π/4+2πk U x=π/4+2πk,k∈z
-5π/2≤-π/4+2πk≤-π /*π/4
-10≤-1+8k≤-4
-9≤8k≤-3
-9/8≤k≤-3/8
k=-1 x=-π/4-2π=-9π/4
-5π/2≤π/4+2πk≤-π /*π/4
-10≤1+8k≤-4
-11≤8k≤-5
-11/8≤k≤-5/8 нет решения
k=-1 x=π/4-2π=-7π/4
ответ {-π/4+2πk;-3π/4+2πk,k∈z};-9π/4;-7π/4
