Дано F(x)=a cos2x+bsin4X F'(7П/12)=4 F'(3П/4)=2 Чему равны a и b??

Question

Дано F(x)=a cos2x+bsin4X F'(7П/12)=4 F'(3П/4)=2 Чему равны a и b??

Answer 1

F(x) = a*cos(2x) + b*sin(4x); 

F'(x)= -2*a*sin(2x) + 4*b*cos(4x);

F'(7П/12)= 4 = -2*a*sin(7П/6) + 4*b*cos(7П/3) = 2*a*sin(П/6) + 4*b*cos(П/3) = a + 2*b;

F'(3П/4)= 2 = -2*a*sin(3П/2) + 4*b*cos(3П) = 2*а - 4*b;

a + 2*b = 4;

a - 2*b = 1;

a = 5/2; b = 3/4;