Question

6 sin ^2+3sinxcosx-cosx ^2=1 помогите пожалуйста

Answer 1

$6 sin ^2x+3sinxcosx-cosx ^2=1;\ 6 sin ^2x+3sinxcosx-cosx ^2=sin^2x+cos^2x;\ 5sin ^2 x+3sinxcosx-2cosx ^2=0; :cos^2x \ 5tg^2 x+3tgx-2=0;\ y=tgx;\ 5y^2+3y-2=0;\ D=3^2-4*5*(-2)=49=7^2;\ y_1=frac{-3-7}{2*5}=-1; y_2=frac{-3+7}{2*5}=0.4;\ tg x=-1; x=-frac{pi}{4}+pi*k;\ tg x=0.4; x=arctg (0.4)+pi*n$

ответ: $-frac{pi}{4}+pi*k;<var></var>$k є Z

$arctg (0.4)+pi*n </var>$, n є Z