решите уравнение
4ctg x/ 1+ ctg^2 x + sin^2 2x+1=0

4сosx/sinx:(1+cos²x/sin²x)+2sin²2x+1=04cosx/sinx*sin²x/(si²x+cos²x)+2sin²2x+14cosx*sinx+2sin²2x+1=04sinxcosx+4sin²xcos²x+1=0(2sinxcosx+1)²=0(sin2x+1)²=0sin2x=-12x=-π/2+2πkx=-π/4+πk/2,k∈z