Question

в)-2cos^2x-5sinx-1=0
г)sin^2x+4sinxcosx-5cos^2=0

Answer 1

в)
$-2cos^2x-5sinx-1=0 \-2(1-sin^2x)-5sinx-1=0 \sinx=y, y in [-1;1] \-2+2y^2-5y-1=0 \2y^2-5y-3=0 \D=25+24=49=7^2 \y_1= frac{5+7}{4} =3 notin[-1;1] \y_2= frac{5-7}{4} =- frac{1}{2} in [-1;1] \sinx=- frac{1}{2} \x_1= -frac{pi}{6} +2pi n, n in Z \x_2=-frac{5pi}{6} +2pi n, n in Z$
г)
$sin^2x+4sinxcosx-5cos^2=0 \ frac{sin^2x}{cos^2x} -4* frac{sinx}{cosx} -5=0 \tg^2x-4tgx-5=0 \tgx=y \y^2-4y-5=0 \D=16+20=36=6^2 \y_1= frac{4+6}{2} =5 \y_2= frac{4-6}{2} =-1 \tgx=5 \x_1=arctg(5)+pi n, n in Z \tgx=-1 \x_2=- frac{pi}{4} +pi n, n in Z$