Question

3sin^2x-4sinx*cosx+cos^2x=3
помогите

Answer 1

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

Answer 2

$3sin^2x-4sinx*cosx+cos^2x=3 \\3sin^2x-4sinx*cosx+cos^2x-3=0 \\-3(1-sin^2x)-4sinx*cosx+cos^2x=0 \\-3cos^2x-4sinx*cosx+cos^2x=0 \\-2cos^2x-4sinx*cosx=0 \\-2cosx(cosx+2sinx)=0 \\1 sluchai:\-2cosx=0\cosx=0\x_1=frac{pi}{2}+pi n ;nin Z \\2sluchai:\cosx+2sinx=0\razdelim pochlenno na cosxneq 0 \1+2tgx=0\tgx=-frac{1}{2}\x_2=arctg(-frac{1}{2})+pi k ;kin Z$