Ц1 решить систему уравнений sin x=y-3,cos x=y-2.

Otvet: 1. x = 2*Pi*k, y = 3 2. x = -Pi/2+2*pi*k, y = 2 cos(x) - sin(x) = 1 2*cos^2(x) - 2*sin(x/2)*cos(x/2) = 2 cos^2(x/2) - sin(x/2)cos(x/2) = 1 a. cos^2(x/2) ne raven 0 1 - tg(x/2) = 1+tg^2(x/2) tg(x/2)*( tg(x/2)+1 ) = 0 tg(x/2) = 0 -> x/2 = Pi*k -> x = 2*Pi*k ; y = 2+cos(x) = 3 (y = 3+sin(x) = 3) b. 1+tg(x/2) = 0 tf(x/2) = -1 x/2 = -Pi/4 + Pi*k x = -Pi/2+2*Pi*k y = 2+cos(x) = 2 (3+sin(x) = 3-1=2)

Синус хельд = уль-3, косинус хельд = уль-2.