Question

1)sin x+2cos x=0
2)4cos x-sin x=0
3)4cos^2 x+0,5 sin^2 x+3sin^2 x=3
4)3sin^2x+4sinx*cosx+ 5 cos^2 x=2
Помогите пожалуйста ;з

Answer 1

$sinx+2cosx=0|:cosx\tgx+2=0\tgx=-2\x=-arctg2+pi n, ; nin Z.\\\4cosx-sinx=0|:(-cosx)\tgx-4=0\tgx=4\x=arctg4+pi n, ; nin Z.\\\4cos^2x+0,5sin^2x+3sin^2x=3\4cos^2x+3,5sin^2x=3(sin^2x+cos^2x)\4cos^2x+3,5sin^2x-3sin^2x-3cos^2x=0\cos^2x+0,5sin^2x=0\1-sin^2x+0,5sin^2x=0\1-frac{1}{2}sin^2x=0\sin^2x=2\sinx=pm sqrt2\\sinx=sqrt2\x=(-1)^narcsin sqrt2 +pi n, ; nin Z;\\sinx=-sqrt2\x=-arcsinsqrt2 +pi n, ; nin Z.$


$3sin^2x+4sinxcosx+5cos^2x=2\3sin^2x+4sinxcosx+5cos^2x=2(sin^2x+cos^2x)\3sin^2x+4sinxcosx+5cos^2x-2sin^2x-2cos^2x=0\sin^2x+4sinxcosx+3cos^2x=0|cos^2x\tg^2x+4tgx+3=0\tgx=u\u^2+4u+3=0\D:16-12=4\u=frac{-4pm 2}{2}=-4pm 1\\u_1-3\tgx=-3\x=-arctg3+pi n, ; nin Z;\\u_2=-5\tgx=-5\x=-arctg5+pi n, ; nin Z.$