Question

sin^2x-5sinx *cosx+cos^2x=-1​

Answer 1

Объяснение:

$\sin^2(x)-5\sin(x)cos(x)+cos^2(x)=-1$

Прим этом

$\sin^2(x)+\cos^2(x)=1$

$\sin(2x)=2\sin(x)\cos(x)$

Получаем:

$1-5(\frac{1}{2} \sin(2x))=-1\\2.5 \sin(2x))=2\\\sin(2x)=\frac{2}{2.5} \\\sin(2x)=\frac{4}{5}=0.8\\\left \{ {{2x = acrsin(0.8)+2\pi n} \atop {2x = \pi-acrsin(0.8)+2\pi n}} \right. \\\left \{ {{x = 0.5acrsin(0.8)+\pi n} \atop {x = 0.5\pi-acrsin(0.8)+\pi n}} \right.$