1) tgx=0
x=πk, k∈Z.
2) 1+cos2x=0
cos2x= -1
2x= π + 2πn, n∈Z
x=π/2 + πn, n∈Z.
Ответ: πk, k∈Z;
π/2 + πn, n∈Z.
7^х+5×7^(х-2)=378;
7^х+5×7^х×7^(-2)=378;
7^×(1+5×7^(-2))=378;
7^х (54/49)=378;
7^х=378:(54/49);
7^х=343;
х=3;
А.xz+xy+2z+2y = ...x(z+y)+2(z+g)<span>
Б.2ab-2ac+3b-3c = ...2</span>a(b-c)+3(b-c)<span>
В.5ax+10ay+bx+2by = ...5</span>a(x+2y)+b(x+2y)<span>
Г.3ac+6bc+7ax+14 bx = ...3</span>c(a+2b)+7x(a+2b)<span>
Д.2ax+2xy-an-yn = ...2</span>x(a+y)-n(a+y<span>)</span>
Sinx(ssinx - 3cosx) = 0
1) sinx=0
x1 = πn, n∈Z
2) sinx - 3cosx = 0 / cosx ≠ 0
tgx - 3 = 0
tgx = 3
x2 = arctg3 + πk, k∈Z