2sin^2x-5sin 2x cos 2x+2cos^2x=0;2tq²2x - 5tq2x +2 =0 ;
tq²2x - (1/2+2)tq2x +1 =0 ;
[ tq2x =1/2 ; tq2x =2 .
[ 2x=arctq(1/2) +πn ; [ 2x=arctq2 +πn , n∈Z .
x₁=(1/2)*arctq(1/2) +(π/2)*n , n∈Z ; x₂=(1/2)*arctq2 +(π/2)*n, n∈Z .
ответ : (1/2)*arctq(1/2) +(π/2)*n , (1/2)*arctq2 +(π/2)*n, n∈Z .
30х^2-15х-12х+6-13=30х^2-6х
(30х^2 сокращаются)
-27х-7=-6х
-21х=-7
х=21/7=1/3
A) = -3
б) = 4
в) = нет основания ( решить нельзя)
г) = -4
X2-10х+21=0
D= 100-84=16
x1= 10-4/2=3
x2= 10+4= 7
5y2+9y-2=0
D=81+40=121
y1= -9-11/10=-2
y2= -9+11/10= 0,2