Question

помогите пожалуйста с математикой.даю 39 пунктов.

Answer 1

$1) 3sinx=2\ sinx=frac{2}{3}\ x=(-1)^narcsinfrac{2}{3}+pi n, nEZ\ \ 2)5sin^2x+3sinxcosx-3cos^2x=2 | :cos^2x\ 5frac{sin^2x}{cos^2x}+3frac{sinxcosx}{cos^2x}-3frac{cos^2x}{cos^2x}=2frac{1}{cos^2x}\ \ 5tg^2x+3tgx-3=2(1+tg^2x)\ 5tg^2x+3tgx-3-2-2tg^2x=0\ 3tg^2x+3tgx-5=0\ D=3^2-4*3*(-5)=9+60=69\ tgx_1=frac{-3+sqrt{69}}{6}\ x_1=arctgfrac{-3+sqrt{69}}{6}+pi n, nEZ\ tgx_2=frac{-3-sqrt{69}}{6}\ x_2=arctgfrac{-3-sqrt{69}}{6}+pi n, nEZ$

 

$3)5sin^2x+sqrt3sinxcosx+6cos^2x=5 | :cos^2x\ 5frac{sin^2x}{cos^2x}+sqrt3frac{sinxcosx}{cos^2x}+6frac{cos^2x}{cos^2x}=5frac{1}{cos^2x}\ \ 5tg^2x+sqrt3tgx+6=5(1+tg^2x)\ 5tg^2x+sqrt3tgx+6-5-5tg^2x=0\ tg^2x+sqrt3tgx+1=0\ D=(sqrt3)^2-4*1*1=3-4=-1\ D<0$

Корней нет.

 

$4) sin^2x=3cos^2x+sin2x\ sin^2x=3cos^2x+2sinxcosx\ sin^2x-3cos^2x-2sinxcosx=0 |:cos^2x\ tg^2x-2tgx-3=0\ D = (-2)^2-4*1*(-3)=4+12=16\ tgx_1=frac{2+4}{2}=3\ x_1=arctg3+pi n, nEZ\ \ tgx_2 = -1\ x_2=arctg(-1)+pi n, nEZ$