Question

Решить уравнение
3sin²x-2sinxcosx =cos²x

Answer 1

$3sin^2x-2sinxcosx =cos^2x \3sin^2x-2sinxcosx-cos^2x=0 \ 3*frac{sin^2x}{cos^2x} -2 frac{sinx}{cosx} -1=0 \3tg^2x-2tgx-1=0 \tgx=y \3y^2-2y-1=0 \D=4+12=16=4^2 \y_1= frac{2+4}{6} =1 \y_2= frac{2-4}{6} =- frac{1}{3} \tgx=1 \x_1= frac{pi}{4} +pi n, n in Z \tgx=- frac{1}{3} \x_2=-arctg( frac{1}{3} )+pi n, n in Z$