A) cos ^2  x + sinx * sin ( 3п/2 + x ) = 1б) x принадлежит [ -3п ; -3п/2 ]

cos^2x-sinxcosx=sin^2x+cos^2xsin^2x+sinxcosx=0sinx(sinx+cosx)=0sinx=0x=Пk  k=-3  x=-3П            k=-2  x=-2Пsinx=-cosxtgx=-1x=-П/4+Пkk=-2x=-9П/4

cos²x+sinx*sin(3π/2+x)=1cos²x+sinx*(-cosx)-cos²x-sin²x=0sin²x-sinxcosx=0 /cos²x≠0tg²x-tgx=0tgx(tgx-1)=0tgx=0⇒x=πntgx=1⇒x=π/4+πnx=-7π/4;-2π;-11π/4;-3π∈[-3π;-3π/2]