Question

Решение обязательно на листочке! ​

Answer 1

$\displaystyle3*(sin^2x+cos^2x)-4sin^2x-2sin(x)*cos(x)=0\\\\3sin^2x+3cos^2x-4sin^2x-2sinx*cosx=0\\\\-sin^2x-2sinx*cosx+3cos^2x=0| :-1\\\\sin^2x+2sinx*cosx-3cos^2x=0|: cos^2x\\\\tg^2x+2tgx-3=0\\\\tgx=t\\\\t^2+2t-3=0\\\\D=4+12=16\\\\t_{1.2}=\frac{-2 \pm 4}{2}\\\\t_1=-3; t_2=1$

$\displaystyle tgx=1; x=\frac{\pi }{4}+\pi n; n \in Z\\\\tgx=-3; x= -arctg 3+\pi n; n \in Z$