Question

3sin^2x+3sinxcosx+2cos^2x=1

 

Answer 1

$3 sin^{2} x+3sinxcosx+2cos ^{2}x = sin^{2}x+ cos^{2} x$
$2sin^{2}x+3sinxcosx+ cos^{2}x=0$
Разделим на $cos^{2}x$
$2 tg^{2}x+3tgx+1=0$
Сделаем замену $y=tgx$
$2 y^{2} +3y+1=0$
$D=9-8=1$
$y_{1}=-2$
$y_{2}=- frac{1}{2}$
$tgx=-2$
$tgx=- frac{1}{2}$
$x_{1} =-arctg2+ pi k$
$x_{2} =-arctg frac{1}{2} + pi k$