Question

sin2xcos2x= - корень из 2 на 2

Answer 1

sin2xcos2x= -√2/2
ОДЗ: x∈R
sin2xcos2x +√2/2=0
sin2xcos2x +√2/2*(cos²2x+sin²2x)=0
sin2xcos2x +√2/2cos²2x+√2/2sin²2x=0(: сos²2x)
tg2x+√2/2+√2/2tg²2x=0
√2/2tg²2x+tg2x+√2/2=0
tg2x=t
√2/2t²+t+√2/2=0
D<0 корней нет.