1-cosx=sinx*sinx/2

1-cos(x)=sin(x)*sin(x/2)

    2sin²x/2 - 2sin²x/2·cosx/2 = 0

    x/2=2пk

    x/2=Пk

    х =  2πn, n∈Z,            x/2 = 2πk, k∈Z

    x/2 = πn, n∈Z,            cosx/2 = 1

     

    x=2Пk

     

  • 2sin^2x/2=2sin^2x/2cosx/2

  • 1-(cos²x/2-sin²x/2)= 2·sinx/2·cosx/2·sinx/2

    sin^2x/2(1-cosx/2)=0

    Объединяя корни, получим, х =  2πn, n∈Z.

    x=4Пk

    sinx/2=0

    2sin²x/2·( 1 - cosx/2) =0

    sinx/2 = 0       или     1 - cosx/2 = 0

    Ответ:  2πn, n∈Z

  •  cos²x/2 + sin²x/2 - cos²x/2 + sin²x/2 = 2sin²x/2·cosx/2

    x=2Пk

    cosx/2=1

                                        х=4πk, k∈Z.