Question

2sin(0,5п+2x)
———————
ctgx+tg(п-x)

П
Х= —
3

Answer 1

$frac{2sinleft(fracpi2+2xright)}{ctgx+tgleft(pi-xright)}=frac{2cos2x}{ctgx-tgx}=2cos2x:left(frac{cos x}{sin x}-frac{sin x}{cos x}right)=2cos2x:left(frac{cos^2x-sin^2x}{sin xcos x}right)=\\=2cos2x:frac{cos2x}{sin xcos x}=2cos2xcdotfrac{sin xcos x}{cos2x}=2sin xcos x=sin2x\\x=fracpi3:\sin2x=sinfrac{2pi}3=frac{sqrt3}2$