Question

30 баллов!!!тригонометрия!!!
sin2x÷sin(пи-x)=√2
промежуток: [-5пи/2;-пи)

Answer 1

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