Решите срочно 1-cosx-sinx/2=0

sin²x/2+cos²x/2-cos²x/2+sin²x/2-sinx/2=02sin²x/2-sinx/2=0sinx/2(2sinx/2 -1)=0sinx/2=0⇒x/2=πn⇒x=2πnsinx/2=1/2⇒x/2=(-1)^n *π/6+πn⇒x=(-1)^n*π/3+2πn

1)  1-cosx-sinx/2=0(1-cosx) - sinx/2 = 02*sin^2(x/2) - sin(x/2) = 0sin(x/2)*(2* sin(x/2) - 1 ) = 01)  sin(x/2) = 0 x/2 = πn, n∈Zx1 = 2πn, n∈Z2)  2* sin(x/2) - 1 = 0sin(x/2) = 1/2x/2 = (-1)^n*arcsin(1/2) + πk, k∈Zx/2 = (-1)^n*(π/6) + πk, k∈Zx2 = (-1)^n*(π/3) + 2πk, k∈Z