Ребята! Помогите решить cos x-1=sin x/2

cosx=cos^2(x/2)-sin^2(x/2) 1=sin^2(x/2)+cos^2(x/2) cosx+1=cos^2(x/2)-sin^2(x/2)-(sin^2(x/2)+cos^2(x/2))=sin(x/2) -2sin^2(x/2)=sin(x/2) a=sin(x/2) -2aˆ2-a=0 a(2a+1)=0 a1=0 a2=-1/2 1) sin(x/2)=a1=0 x/2=Pi/2+PiN x=Pi+2PiN 2)sin(x/2)=a2=-1/2 x/2=-7Pi/6, 11Pi/6 x=7Pi/3, 11Pi/3

[latex]cosx-1=sin \frac{x}{2} [/latex] [latex]-(1-cosx)=sin \frac{x}{2} [/latex] [latex][sin^2 \frac{x}{2} = \frac{1-cosx}{2}] [/latex] [latex]-2sin^2 \frac{x}{2} -sin \frac{x}{2} =0[/latex] [latex]2sin^2 \frac{x}{2}+sin \frac{x}{2} =0[/latex] [latex]sin \frac{x}{2}(2sin \frac{x}{2}+1) =0[/latex] 1) [latex]sin \frac{x}{2} =0[/latex]  [latex] \frac{x}{2} = \pi k,[/latex] [latex]k[/latex] ∈ [latex]Z[/latex] [latex]x=2 \pi k,[/latex] [latex]k[/latex] ∈ [latex]Z[/latex] 2) [latex]2sin \frac{x}{2}+1 =0[/latex] [latex]2sin \frac{x}{2}=-1[/latex] [latex]sin \frac{x}{2}=- \frac{1}{2} [/latex] [latex] \frac{x}{2} =(-1)^{n+1} \frac{ \pi }{6} + \pi n,[/latex] [latex]n[/latex] ∈ [latex]Z[/latex] [latex]x} =(-1)^{n+1} \frac{ \pi }{3} + 2\pi n,[/latex] [latex]n[/latex] ∈ [latex]Z[/latex]