3sin^2x-4sinx*cosx+cos^2x=3помогите

3-3cos^2x-4sinxcosx+cos^2x=3 -2cos^2x-4sinxcosx=0 cos^2x+2sinxcosx=0 cosx(cosx+2sinx)=0 cosx=0 x=П/2(2l+1) cosx=-2sinx tgx=-1/2 x=П-arctg(0,5)+Пk

[latex]3sin^2x-4sinx*cosx+cos^2x=3 \\\\3sin^2x-4sinx*cosx+cos^2x-3=0 \\\\-3(1-sin^2x)-4sinx*cosx+cos^2x=0 \\\\-3cos^2x-4sinx*cosx+cos^2x=0 \\\\-2cos^2x-4sinx*cosx=0 \\\\-2cosx(cosx+2sinx)=0 \\\\1 sluchai:\\-2cosx=0\\cosx=0\\x_1=\frac{\pi}{2}+\pi n\ ;n\in Z \\\\2sluchai:\\cosx+2sinx=0\\razdelim\ pochlenno\ na\ cosx\neq 0 \\1+2tgx=0\\tgx=-\frac{1}{2}\\x_2=arctg(-\frac{1}{2})+\pi k\ ;k\in Z [/latex]