Question

8-4sin^2х=sin2x * ctgx - 9cosx
sinх не равен 0
х не равен Пk

Answer 1

$8-4sin^2x=sin2xcdot ctgx-9cos x\ D(f): ctgx: xneqpi k, kin Z\ 8-4sin^2x=2sin xcos xcdotfrac{cos x}{sin x}-9cos x\ 8-4sin^2x=2cos^2x-9cosx;\ 8-4+4cos^2x=2cos^2x-9cos x;\ 4cos^2x-2cos^2x+9cos x+8-4=0;\ 2cos^2x+9cos x+4=0;\ cos x=t; -1leq tleq1;\ 2t^2+9t+4=0;\ D=9^2-4cdot2cdot4=81-32=49=(pm7)^2;\ t_1=frac{-9-7}{2cdot2}=frac{-16}{4}=-4notin D(f);\ t_2=frac{-9+8}{2cdot2}=frac{-2}{4}=-frac12tin D(f);\ cos x=-frac12;$
$x=pi-frac{pi}{3}+2pi n, nin Z$