(cos pi/4*cos 2x + sin pi/4*sin 2x) - √2sinx = √2(sin2x + 1)
2(√2/2*cos 2x + √2/2 *sin2x) - √2sinx = √2(sin 2x +1);
2*√2/2 (cos 2x + sin 2x) - √2sinx = √2(sin 2x +1);
cos2x + sin 2x -sinx = sin 2x +1;
1 - 2sin^2 x - sinx = 1;
-sinx(2sinx+1)=0;
1) sinx = 0; x1=pin, n---Z.
2) 2sinx+1=0; sinx= -1/2; x2=(-1)^n(-pi/6)+pin, n---Z.
4х+у=10
х+3у=-3
4х+у=10
х=-3-3у
4(-3-3у)+у=10
-12-12у+у=10
-11у=10+12
-11у=22
у=-2
Х=-3-3*(-2)=-3+6=3
<span>Ответ:3;-2</span>
<span>A(0,1,2,3) и B(-1,2,3,4,5,6)
A</span>∩B={2;3}