Решите уравнение : sin(-x/2)=1/2 2cos(2x+pi/4)=-sqrt2 , умоляю

Sin(-x/2)=1/2
-sinx/2=1/2
sinx/2=-1/2
x/2=(-1)^(n+1)π/6+πn, n∈Z
x=(-1)^(n+1)π/3+2πn, n∈Z

2cos(2x+π/4)=-√2
cos(2x+π/4)=-√2/2
2x+π/4=+/-(π-π/4)+2πk, k∈Z
2x+π/4=+/-3π/4+2πk, k∈Z
2x=+/-3π/4-π/4+2πk, k∈Z
2x=3π/4-π/4+2πk=π/2+2πk, k∈Z      x1=π/4+πk, k∈Z
2x=-3π/-π/4+2πk=-π+2πk, k∈Z        x2=-π/2+πk, k∈Z