Question

Sin^2 x-5ctgx=1+2sin2x

Answer 1

$sin^2x-5ctgx=1+2sin2x \ sin^2x-5ctgx=sin^2x+cos^2x+2sin2x \ -5cos x=cos^2xcdot sin x+4sin^2xcos x \ cos x(cos xsin x+9sin^2 x+5cos^2x)=0 \ cos x=0 \ x= frac{pi}{2} +pi n,n in Z$
$9sin^2x+sin xcos x+5cos^2x =0|:cos^2 x \ 9tg^2x+tg x+5=0 \ D=1-9*5*4<0$

Ответ: $frac{pi}{2} +pi n$